Algebra Class 10 (zambak)

y

A function from a set D to a set R is a rule that maps each element of D to a single element of R.

y = f(x)

b

f: D → R ↓ ↓

f: D ® R f: x ® y= f(x)

domain range O For each x in D there exists a single element y in R such that f(x) = y. x is called the variable of f and y = f(x) is called the image of x. The set of images of all the elements of D is called the image set of f.

x

a f(a) = b b is the image of a

Consider the function f: A → B, f(x) = x2. We can write this function in different ways: f(x) = x2, y = x2 or f: x → x2. All of these mean the same function. A is the domain and B is the range of f.

A. DOMAIN AND RANGE OF A FUNCTION Unless stated otherwise, the domain of a function f(x) is the largest set of real x-values for which f(x) is defined. The range of the function is a set which includes at least all the images of the elements in its domain. The domain and range of many functions are subsets of the set of real numbers. The largest possible range of a function is \. Let us look at the domain and range of some common types of function. In these examples we will use the letter D to mean the domain of the function. The notation D( f ) means the domain of the function f. Type of function

Form

polynomial

f(x) = anxn+an–1xn–1+...+a0

function

n Î ¢+ È {0}

Domain ¡

Examples f(x) = 2x+5

D( f ): ¡

f(x) = 2x2–3x+1

D( f ): ¡

As stated in the table, the domain of any polynomial function f is \. The range of the polynomial function depends on the function itself. For example, let us draw the graph of the function y = f(x) = xn and find its domain and range when n is odd or even. y when n is even

n

y = f(x) = x

when n is odd O

x

f: ¡ ® ¡+ Ç {0} ¯ ¯

domain range

10

y

y = f(x) = xn

x

O

f: ¡ ® ¡ ¯ ¯

domain range

Algebra 10

Type of function

Form

rational function

f(x) =

Domain

g(x)

Examples f(x) =

2x – 3 x+1

f(x) =

x2 + 5 D( f ): ¡ – {5, –2} (x – 5)(x + 2)

¡–{x | h(x)=0}

h(x)

D( f ): ¡ – {–1}

The value of the denominator in a rational expression cannot be zero, so any numbers which make the denominator zero must be excluded from the domain of a rational function. 1 As an example, let us look at the graph of the function y = x– n = n . x y

y

y = f(x) =

when n is even

1 xn

when n is odd x

O

f: ¡ – {0} ® ¡+

y = f(x) =

1 xn x

O

f: ¡ – {0} ® ¡ – {0}

We can see that the domain of a rational function changes according to the function. Type of function

Form

Domain

radical function

f(x) = n g(x) n is an odd integer

¡

When the index of a radical expression is odd, the radicand can be negative, positive or zero. Therefore there is no restriction on the value of x and so the domain is \. The figure at the right shows the graph of y = n x when n is odd and n > 1. We can see that the range of the function is the set of real numbers. Functions

Example f(x) = 3 x2 + 5x

D( f ): ¡

y y = f(x) = n x

x

O f: ¡ ® ¡ n is odd, n > 1

11

Type of function

Form

Domain

Examples

f(x) = n g(x) f(x) = n is an even ¡ – {x | g(x) < 0} f(x) = integer

radical function

x2 – 2 4

When the index of a radical expression is even, the radicand cannot be negative, so we must exclude any numbers which make the radicand negative from the domain.

range of y = x are the set \ ∪ {0}. n

EXAMPLE

1

+

nx y = f(x) = ñ

f : [0, +¥) ® [0, +¥) x

n is even

Domain

Range

f(x) = x2 + 2x + 1

¡

[0, +¥)

f(x) = x2 – 2x – 3

¡

f(x) = x3 – x2 + x + 1 f(x) =

1 x+1

f(x) =

x3 – 1 x – 5x + 6 2

f(x) = 1 – x2

12

O

The following table shows some more examples of the domain and range of different functions. Write the missing values in the table. Function

Solution

x – 3x + 2 D( f ): ¡ – (1, 2)

y

As we can see in the graph opposite, since the radicand is non-negative both the domain and

D( f ): ¡ – (–ñ2, ñ2)

2

¡ (–¥, 0) È (0, +¥) ¡ – {2, 3} [0, 1]

The solution is left as an exercise for you. Algebra 10

EXAMPLE

2

Solution

Find the domain and range of the function f ( x) = x2 +5 x +6. We know that the radicand of a square root function cannot be negative. Let us look at the sign of the radicand x2 + 5x + 6: x 2

x + 5x + 6

–3 +

–2 –

+

The radicand is non-negative in the intervals (–∞, –3] and [–2, +∞). Therefore the domain of the function is (–∞, –3] ∪ [–2, +∞). As x increases, the value of y also increases without limit. So the range is [0, +∞). In conclusion, the domain of f is (–∞, –3] ∪ [–2, +∞) and the range is [0, +∞).

Note If a function f is a sum or difference of different functions then the domain of f is the intersection of the domains of each function.

EXAMPLE

5

Solution

The image set I of a function f: A → B is the set of images of all the elements of A, so I ⊆ B.

Find the image set of each function over the given interval. a. f(x) = 3x + 6, x ∈ [0, +∞)

b. f(x) = x2 – 2x + 8, x ∈ [–1, 2]

c. f(x) = x2 – 4x – 5, x ∈ [–1, 1]

d. f(x) = 2x – x2, x ∈ [0, 3]

We can find each range by drawing the graph of the function over the given interval. a. As we can see in the graph, f(x) is an increasing function. The solid line shows the graph on the interval x ∈ [0, +∞). On this interval the minimum value of f is f(0) = 6, and the maximum value goes to infinity.

y y = f(x) = 3x + 6 6

So the image set of f on this interval is [6, +∞). –2 O

Functions

min value

x

13

b. The figure shows the graph of the function y = f(x) = x2 – 2x + 8. The solid line shows

y

the graph on the interval x ∈ [–1, 2]. Since the extremum (vertex) of a function b b f(x) = ax2 + bx + c is (– , f (– )), 2a 2a the minimum value of the function on this −2 b interval is f ( − ) = f ( − ) = f (1) = 7. 2a 2

y = x2 – 2x + 8 11

max value 8 7

min value

Its maximum value on this interval is f(–1) = 11. So the image set of f on this interval is [7, 11].

c. The figure shows the graph of y = f(x) = x2 – 4x – 5. The solid line shows the graph on the interval x ∈ [–1, 1]. On this interval, min f(x) = f(1) = –8 and max f(x) = f(–1) = 0.

–1 O

y

1 2 3 4

14

x

–5

d. The solid line in the figure shows the graph of y = f(x) = 2x – x2 on the interval

So the image set of the function on this interval is [–3, 1].

5

–2 –1 O

min value

On this interval, min f(x) = f(3) = –3 and max f(x) = f(1) = 1.

y = x2 – 4x – 5

max value

So the image set of f on this interval is [–8, 0].

x ∈ [0, 3].

x

1 2

–8 –9

y max value 1 3 O

1

2

x

–1 –2 –3

min value

y = 2x – x2

Algebra 10

EXAMPLE

6

Find the range of each function for its largest domain. a. f ( x) = − x2 + 4x +5

Solution

a. Let us define g(x) = –x2 + 4x + 5 and plot its graph (shown opposite).

y 9

As we can see, the function is positive over the interval x ∈ [–1, 5]. So f ( x) = − x2 + 4x +5 is defined on the interval x ∈ [–1, 5]. Also, on this interval the minimum value of f(x) is 0 and the maximum value of f(x) is ñ9, so we can write 0≤

− x2 + 4x +5 ≤ ñ9

0≤

− x2 + 4x +5 ≤ 3

g(x) ³ 0

5

–1 O

2

5

x

y = g(x) = –x2 + 4x + 5

0 ≤ f(x) ≤ 3. So the range of f is [0, 3].

EXAMPLE

8

Solution

Find the domain of f ( x) = | x − 1| − | x + 2|.

The radicand |x – 1| – |x + 2| must be non-negative, i.e. |x – 1| – |x + 2| ≥ 0. This gives |x – 1| ≥ |x + 2|. Taking the square of both sides gives us x2 – 2x + 1 ≥ x2 + 4x + 4 –6x ≥ 3 1 1 x ≤ – . So the domain of the function is (–∞, – ]. 2 2

EXAMPLE

9

Solution

2 Find the domain of f ( x) = 3 − 12 − x .

We have the radicands 12 – x2 and 3 − 12 − x2 , and both of them must be non-negative: 12 – x2 ≥ 0

and

3 − 12 − x2 ≥ 0

9 ≥ 12 – x2 x2 – 3 ≥ 0. Functions

15

Let us solve each quadratic inequality by constructing its sign table: x

–2ñ3 +

–

12 – x2

x

2ñ3 –

+

x2 – 3 –2ñ3

–ñ3

O

ñ3

–ñ3

ñ3

–

+

2ñ3

The domain of the combined function f ( x) = 3 – 12 – x2 is the intersection of these intervals: D( f ) = [–2ñ3, –ñ3] ∪ [ñ3, 2ñ3].

Check Yourself 1 1. Find the domain of each function. a. f ( x) =

x x +2

b. f ( x) = 3 2 x +1+ x2 – 1

2

c. f ( x) = 2 x − 4 x +1 x −1

d. f ( x) = 2 x − 1 − 3x +1 2. Find the image of each function over the given interval. a. f(x) = 2x + 1, x ∈ [1, 5)

b. f(x) = x2 – 4x – 5, x ∈ [1, 3)

c. f ( x) = x2 + 2, x ∈ ( 2, 7] 3. Find the range of each function for its largest domain. 2 a. f ( x) = − x +7 x − 12

Answers 1. a. \

b. \ – (–1, 1)

2. a. [3, 11) 3. a. [0, 16

b. [–9, –8]

c. \ – (–1, 1]

7 d. [ , ∞) 4

f. (2, 3]

1 ] 2 Algebra 10

B. COMPOSITE FUNCTION composite function

Definition

A function that is formed by the composition of two or more elementary functions is called a composite function. x ⎯→

f

f(x)

⎯→

g

⎯→ g( f(x))

The function g(f(x)) is called the composite of g with f. The composite function g(f(x)) is also sometimes written as (g ο f )(x). The composition of functions is associative but not commutative, i.e. (f ο (g ο h))(x) = ((f ο g) ο h) but (f ο g)(x) ≠ (g ο f )(x).

EXAMPLE

10

Solution

Given f(x) = x2, g(x) = sin x and h(x) = 2x + 5, find a. f(g(x)).

b. g(f(x)).

c. f(h(g(x))).

a. f(g(x)) = f(sin x) = sin2 x b. g( f(x)) = g(x2) = sin x2 c. f(h(g(x))) = f(h(sin x)) = f(2sin x + 5) = (2sin x + 5)2

EXAMPLE

11

Solution

Write each function as a composite function. a. f(x) = (3x – 5)3 a. Notice that these are not the only possible solutions to the question. We could have chosen different elementary functions and still achieved the same function. For example, for the function f(x) = (3x – 5)3, we could have chosen g(x) = 3x, h(x) = x – 5 and t(x) = x3 to get f(x) =(3x – 5)3 = t(h(g(x))).

Check Yourself 2 Write each function as a composite function. 3 7 ⎛ x +5 ⎞ 2x − 1 ⎞ 1. f ( x) = ⎛⎜ 2. f ( x) = sin ⎜ 2 ⎟ ⎟ ⎝ x +1 ⎠ ⎝ 7 ⎠ Functions

17

C. INVERSE OF A FUNCTION Recall the definition of inverse function: if the function f: D → R is both a one-to-one function and an onto function then the function f –1: R → D is called the inverse of f. f(x) = y ⇔ f –1(y) = x A function f: A → B is a one-to-one function if for each x1 ≠ x2 in A, f(x1) ≠ f(x2).

To find the inverse of a given function y = f(x) it is enough to find x in terms of the variable y.

y y = f(x) = 2x – 1 y=x

For example, let us find the inverse of the polynomial function f : \ → \, f (x)= 2x – 1:

y = f –1(x) =

Write y = f(x): y = 2x – 1. A function f: A → B is an onto function if for any y ∈ B there exists an x ∈ A such that f(x) = y.

Express x in terms of y: x =

1

y +1 = f −1( y). 2 –1

Finally, express the inverse function in terms

O

x+1 2 x

1

–1

of the variable x: f −1( x) = x +1. 2 This is the inverse of f(x) = 2x – 1.

Recall that the graph of a function and the graph of its inverse are symmetric with respect to the line y = x.

Let us recall the inverse of some common types of function: Function

Form

linear function

f(x) = ax + b

rational function

f (x) =

ax + b cx + d

Inverse –1 f (x) =

x–b a

f –1(x) =

– dx + b cx – a

Look at some more examples of inverse functions: Function 3x + 1 2

f: ¡ ® ¡,

f(x) =

f: ¡ ® ¡,

f(x) = 2 – 3x

f: ¡ – {1} ® ¡ – {2}, f(x) = 18

Inverse

2x + 1 x–1

f –1: ¡ ® ¡, f –1(x) =

2x – 1 3

–1 –1 f : ¡ ® ¡, f (x) =

–x + 2 3

f –1: ¡ – {2} ® ¡ – {1}, f –1(x) =

x+1 x–2 Algebra 10

EXAMPLE

12

Solution

Find the inverse of the function f : [2, ∞) → [0, ∞), f (x) = x2 – 4x + 4. In the given domain and range, f(x) is both one-to-one and onto, so its inverse is a function. We can find this inverse function as follows: x2 – 4x + 4 (x – 2)2 x–2 x

= = = =

y y = f(x) = x2 – 4x + 4

y y ñy ñy + 2 = f –1(y).

4

So the inverse is

O

f –1: [0, ∞) → [2, ∞), f –1(x) = ñx + 2. EXAMPLE

13

Solution

x

2

The function f: \ → \, f(x) = x3 – 3x2 + 4x – 1 is given. Find the real number a which satisfies the equation f(a) = f –1(a). To solve the problem we have to find the intersection of the graphs of f and f –1. However, we know that the graph of a function and the graph of its inverse are symmetric with respect to the line y = x. In other words, the intersection of the two graphs will be on this line. At the intersection point, therefore, y = f(x) = x, and so f(a) = a = f –1(a), as shown at the right. If f(a) = a then a3 – 3a2 + 4a – 1 a3 – 3a2 + 3a – 1 (a – 1)3 a

= = = =

y y = f(x) y = x a

–1

y = f (x) O

a

x

a 0 0 1.

Check Yourself 3 1. Find the inverse of each function. x a. f : \ → \, f ( x) = 3 − 2 5x − 2 ⎧5 ⎫ b. f : \ − {1} → \ − ⎨ ⎬ , f ( x) = 2x − 2 ⎩2 ⎭

c. f : \ → \, f(x) = x3 – 3x2 + 3x 2. For each function, find the real number a which satisfies the equation f(a) = f –1(a). b. (x) = 8x3 – 12x2 + 7x – 1

a. f(x) = 3x + 1 Answers 1. a. 6 – 2x Functions

b.

2x – 2 2x – 5

c.

3

x – 1 +1

2. a. –

1 2

b.

1 2 19

D. CONSTANT, INCREASING AND DECREASING FUNCTIONS Let f: D → R be a function and let I ⊂ D and x1, x2 ∈ I and such that x1 < x2. 1. If f(x1) = f(x2) for all x1, x2 ∈ I then f is called a constant function on the interval I. We write a constant function as f(x) = c (c ∈ \). If f is a constant function, x1 < x2 ⇔ f(x1) = f(x2) = c.

y c

x1

A

y=c

x2

O

f(x) is constant on I = [AB]

2. If f(x1) < f(x2) for all x1, x2 ∈ I then f is called an increasing function on the interval I. If x1 < x2 and f(x1) ≤ f(x2) then f is decreasing called a non-d function.

x

B

y

If f is an increasing function,

f(x2)

x1 < x2 ⇔ f(x1) < f(x2).

f(x1) O

A

y = f(x)

x1

x2 B

x

f(x) increases on I = [AB]

Note The increasing function f: \ → \, f(x) = x is called the identity function.

3. If f(x1) > f(x2) for all x1, x2 ∈ I then f is called a decreasing function on the interval I.

y= f(x) y

If f is a decreasing function, x1 < x2 ⇔ f(x1) > f(x2). If x1 < x2 and f(x1) ≥ f(x2) then f is called a non-iincreasing function.

EXAMPLE

14

Solution

f(x1) f(x2) A

O

x1

x2 B

x

f(x) decreases on I = [AB]

Given that f(x) = (5 – a)x2 + (b + 2)x – 3 is a constant function, find a and b. Since f is a constant function, the coefficients of x2 and x must be zero: 5 – a = 0 and b + 2 = 0. So a = 5 and b = –2.

20

Algebra 10

EXAMPLE

15

Solution

Show that a. f: [0, ∞) → \, f(x) = ñx + x3 is an increasing function. 1 b. f: \ – {0} → \, f ( x) = is a decreasing function. x a. Let x1, x2 ∈ [0, ∞) such that x1 < x2. For all x1 < x2, x1 < x2 and x13 < x23. So x1 + x13 < x2 + x2 3 and f(x1) < f(x2). So f is an increasing function on [0, ∞). b. First let x1, x2 ∈ (0, ∞) such that x1 < x2. For all x1 < x2 , Now let x1, x2 ∈ (–∞, 0) such that x1 < x2. For all x1 < x2 ,

1 1 > and so f(x1) > f(x2). x1 x2 1 1 > and so f(x1) > f(x2). x1 x2

In both cases f is decreasing, so f is a decreasing function on \ – {0}. EXAMPLE

16

Solution

Find the interval(s) on which each function decreases and/or increases. 1 a. y = 1 – 2x b. y = x2 – 3x c. y = 2 x We can draw a graph of each function to determine the intervals. a. We can see from the graph that f(x) = 1 – 2x decreases on (–∞, ∞).

y y = f(x) = 1 – 2x

1

b. f decreases on the interval ( −∞,

x

1 2

O

y

3 ]. 2

y = f(x) = x2 – 3x

3 f increases on the interval [ ,+∞). 2

3 2 O

3

x

9 –4

Functions

21

c. f increases on the interval (–∞, 0).

y y = f(x) =

f decreases on the interval (0, +∞).

O

EXAMPLE

17

Solution

Find the value a + b if f ( x) =

1 x2

x

( a − 2)x2 + bx + 4 is a constant function. 3x + 2

( a − 2)x2 + bx + 4 = k ( k ∈ \). 3x + 2 This gives (a – 2)x2 + bx + 4 = 3kx + 2k.

Since f is constant, we can write f ( x) =

By the equality of polynomials, we can write (a – 2) = 0 b = 3k 4 = 2k, which gives a = 2, k = 2, b = 6. So a + b = 8.

EXAMPLE

18

The function f ( x) =

( a − 1)x2 +( b +5)x is an identity function. Find a and b. 2 x +1

Solution Since f is an identity function, f(x) = x. So ( a − 1)x2 +( b +5) x = x, i.e. 2 x +1 ( a − 1)x2 +( b +5) x = 2 x2 + x.

By the equality of polynomials, a – 1 = 2 and b + 5 = 1. So a = 3 and b = –4.

Check Yourself 4 1. f ( x) =

3mx +1 is a constant function. Find m. 6 x +5

2. f(x) = (m – 2)x + n + 5 is an identity function. Find m + n. 22

Algebra 10

3. Decide whether each function increases or decreases on the given interval. a. f(x) = 2x + 1, x ∈ \ b. f(x) = 1 – x, x ∈ \ c. f(x) = –x2 – 8x + 1, x ∈ (–4, ∞) d. f(x) = –x2 – 2x – 1, x ∈ (–∞, 1) e. f(x) = x3 + 1, x ∈ \ Answers 1. 2 5

2. –2

E. EVEN AND ODD FUNCTIONS Definition

even function, odd function Let f: D → R be a function. 1. If f(–x) = f(x) for all x ∈ D then f is called an even function. 2. If f(–x) = –f(x) for all x ∈ D then f is called an odd function. For example, the cosine function is an even function because cos (–x) = cos x. Similarly, the sine, tangent and cotangent functions are odd functions because sin(–x) = –sinx, tan (–x) = –tan x and cot (–x) = –cot x. The graph of an even function is symmetric with respect to the y-axis. y

Not all functions are even or odd. For example, f(x) = x + 1 is neither even nor odd.

y y = f(x) = cos x

1 – –2p

3p 2

p –p

p O – 2

p 2

2p 3p 2

x

O

–1 f(–x) = f(x) cos (–x) = cos (x)

Functions

y = f(x) = x2

x

f(–x) = f(x) Even functions

(–x)2 = (x)2

23

The graph of an odd function is symmetric with respect to the origin. y

y The following rules help us to calculate the parity (even or odd) of the sum and product of even functions (E) and odd functions (O): E±E=E O±O=O E ± O = neither E nor O E⋅E=E E⋅O=O O ⋅ O = E.

EXAMPLE

19

1 –2p

–p

y = f(x) = sin x

O

p

2p

x

–

3p 2 –

–1

p 2

O

p 2

3p 2

f(–x) = –f(x)

f(–x) = –f(x)

sin (–x) = –sin (x)

tan (–x) = –tan (x)

x

y y = f(x) = x3

x

O

f(–x) = –f(x) (–x)3 = –x3

Odd functions

Determine whether each function is even, odd, or neither even nor odd. a. f(x) = x2 + 3x + 2

b. f(x) = 7 tanx + x3

e. f ( x) = x2 − 6 x +9 + x2 +6 x +9 Solution

y = f(x) = tan x

2 c. f ( x) = x6 – 28 x +x

d. f(x) = 3x + 3–x

f. f(x) = |x – 3| + |x + 4| g. f(x) = –3x2 + 2|x| – 5

Let us find f(–x) and compare it with f(x) in each case. a. f(–x) = x2 – 3x + 2, so f is neither even nor odd. b. f(–x) = –7 tanx – x3 = –(7tan x + x3) = –f(x), so f is odd. x2 – 2 = f ( x) , so f is even. c. f (– x) = 6 x + x8 d. f(–x) = 3–x + 3x = 3x + 3–x = f(x), so f is even. e. f (– x) = x2 +6 x +9 + x2 – 6 x +9 = f ( x) , so f is even. f. f(–x) = |–x – 3| + |–x + 4|, so f is neither even nor odd. g. f(–x) = –3x2 + 2|x| – 5 = f(x), so f is even.

EXAMPLE

20

f: \ → \, f(x) is an odd function such that f(–2) = k + 5 and f(2) = 2k + 3. Find k.

Solution Since f is an odd function, f(–x) = –f(x) and f(–2) = –f(2). So

k + 5 = –(2k + 3) k + 5 = –2k – 3 3k = –8 8 k= – . 3

24

Algebra 10

EXAMPLE

21

y

Complete the graph of the function if it is a. even.

b. odd. x

O

Solution

a. The graph of an even function is symmetric with respect to the y-axis. y

b. The graph of an odd function is symmetric with respect to the origin. y

y = f(x)

O

x

x

O

y = f(x) f(x) is an even function

f(x) is an odd function

Check Yourself 5 1. Determine whether each function is even, odd, or neither even nor odd. a. f(x) = |x| + cos x

b. f(x) = x3 + sin x

d. f(x) = cos x4 – x3 sin x

e. f ( x) =

x cos( x3 )

c. f(x) = x4 + x2 + 1 sin x + tan x

f. f ( x) = 2

x3

2. f is an odd function and g is an even function. f(–2) + g(1) = 8 and g(–1) + f(2) = 6 are given. Find f(–2) and g(–1). Answers 1. a. even b. odd c. even d. even e. odd f. even

Functions

2. f(–2) = 1, g(–1) = 7

25

EXERCISES

1 .1

A. Domain and Range of a Function

4. Find the range of each function for its largest

1. State the domain and range of each function. y

a.

y = f(x)

y

b.

y = g(x)

5

O

a. f ( x) = – x2 + 4 b. f ( x) = – x2 – 10 x – 9

x

x

O

B. Composite Function y

c.

domain.

d. 3 2

y = h(x) –4 –3

3

–2 –1

x

O

5. Given f(x) = ñx, g(x) = x2, and h(x) = x + 1,

y

write each function. y = t(x)

11 O –1 –2

a. g(h(f(x)))

b. f(h(g(x)))

x

6. Write each function as a composite of elementary functions. a. f ( x) = 5 –

1 x +1

2. Find the domain of each function. a. f ( x) = 9 – | x2 – 4| b. f ( x) =

C. Inverse of a Function

2

x – 3x – 4 x2 – 1

7. Find the inverse of each function.

2x c. f ( x) = 3x +1+ x–5 2

a. f: \ → \, f ( x) =

2−x 5

–2 x – 3 x+ 3 c. f: [1, +∞) → [1, +∞), f(x) = 4x2 – 8x + 5

b. f: \ – {3} → \ – {2}, f ( x) =

3. Find the range of each function over the given interval.

8. For each function, find the real number a

a. f(x) = 1 – 3x,

x ∈ [–2, 4)

which satisfies the equation f(a) = f –1(a).

b. f(x) = x2 – 2x – 3,

x ∈ (2, 4]

a. f(x) = 5x – 2

2

c. f(x) = –x + 4x + 5, x ∈ [0, 1) 26

b. f(x) = x3 – 6x2 + 13x – 8 Algebra 10

D.Constant, Increasing and Decreasing Functions 9. f ( x) =

ax2 + 2 x + b is a constant function. Find 3x2 + bx + 2b

a and b.

10. Determine whether each function is increasing or decreasing on the given interval. a. f(x) = x2 – 6x + 1,

x ∈ (3, ∞)

b. f(x) = –x2 + 4x – 3,

x ∈ (–∞, 2)

c. f(x) = –x3 + 3,

x∈\

11. The function f(x) = 2x2 – 7x – 15 is given. a. On which interval does the function decrease? b. On which interval does the function increase?

E. Even and Odd Functions 12. Determine whether each function is even, odd, or neither even nor odd. a. f(x) = x5 + x3 + x b. f ( x) =

cos x + x2 3+ x4

c. f ( x) = (

x ⋅ tan x 5 ) x3 + sin x

d. f(x) = sin (tan(x3 + x)) e. f(x) = x4 ⋅ sin x Functions

27

A. PIECEWISE FUNCTION piecewise function A function that is defined by different formulas on different intervals of its domain is called a piecewise function.

Definition

EXAMPLE

22

The piecewise function f :

→

⎧2 x + 1 if x > 2 ⎪ ⎪ , f ( x) = ⎨x2 if 0 ≤ x ≤ 2 is given. ⎪ ⎪⎩ – x2 – 1 if x < 0

a. Draw the graph of f. Solution

b. Find f(–5) + f(2) + f(3).

a. When x > 2, we draw the graph of y = 2x + 1,

y = f(x)

y

when 0 ≤ x ≤ 2, we draw the graph of y = x and 2

5

when x < 0, we draw the graph of y = –x2 – 1.

4 3 2 1

b. When x = –5, f(x) = –x2 – 1. So f(–5) = –(–5)2 – 1 = – 26. When x = 2, f(x) = x2. So f(2) = 22 = 4.

O

When x = 3, f(x) = 2x + 1. So f(3) = 2 ⋅ 3 + 1 = 7.

–1

Hence, f(–5) + f(2) + f(3) = –26 + 4 + 7 = –15. y = –x2 – 1

EXAMPLE

23

Solution

The domain of the function f(x) shown in the figure is [0, 3]. Define f(x) as a piecewise function. The graph consists of three line segments. Working from left to right: 1 the first line segment is valid for 0 ≤ x < 1 and f ( x) = , 2 the second segment is valid for 1 ≤ x ≤ 2 and f(x) = 1,

x

2

1

y = x2

y = 2x + 1

y y = f(x)

2 1 0.5 O

1

2

3

x

and the third segment is valid for 2 < x ≤ 3 and f(x) = 2. So the definition is ⎧1 ⎪ 2 if 0 ≤ x < 1 ⎪ f ( x) = ⎨ 1 if 1 ≤ x ≤ 2 ⎪ ⎪ ⎩2 if 2 < x ≤ 3. 28

Algebra 10

Check Yourself 6 1. The piecewise function f : Calculate f(f(f(–1))).

⎧ x2 if x > 3 ⎪ → , f ( x) = ⎨3 x + 4 if 0 ≤ x ≤ 3 is given. ⎪ x3 + 2 if x < 0 ⎩

2. Sketch the graph of each piecewise function. a. f :

⎧⎪ x if x ≥ 1 → , f ( x) = ⎨ ⎪⎩ − x if x < 1

b. f :

⎧⎪ x2 if x > 1 → , f ( x) = ⎨ 2 ⎩⎪ − x if x ≤ 1

c. f :

⎧ 2 x − 1 if x < 0 ⎪⎪ → , f ( x) = ⎨ 1 if 0 ≤ x < 1 ⎪ ⎪⎩ 1 − x if x ≥ 1

Answers 1. 49

B. ABSOLUTE VALUE FUNCTION Recall that for any number x, the absolute value of x (written |x|) is the distance between x and the origin on a number line.

y

⎧ x if x ≥ 0 |x| = ⎨ ⎩ − x if x < 0

y = |x|

x

O

absolute value function

Definition

The absolute value function |f(x)| is defined as ⎧ f ( x) if f ( x) ≥ 0 f ( x) = ⎨ ⎩− f ( x) if f ( x) < 0. EXAMPLE

24

Solution

Functions

Draw the graph y = |x – 3|. We begin by drawing the graph y = x – 3. We then draw the graph y = |x – 3| by reflecting the negative part of the graph.

y

y

3 O –3

x

3

O y=x–3

3

x

y = |x – 3|

29

Note When solving absolute value equations or inequalities or when drawing the graph of an absolute value function, begin by finding the intervals in which the value of the function is negative, positive or zero. EXAMPLE

25

Solution 1

Draw the graph y = |x2 – 1|. x

Let us construct the sign table for x2 – 1.

–1

Solution

y = x2 –1

O

x

1

y = –x2 + 1 y

y = x2 – 1

y

y = |x2 – 1|

1

method: first we graph y = x2 – 1 and then we reflect the negative y-values in the graph with respect to the x-axis.

26

y

+

1

⎧ x2 – 1 if x ≤ –1 or x ≥ 1 So|x – 1| = ⎨ 2 ⎩ − x +1 if – 1 < x < 1. Now we can draw the graph, shown opposite.

EXAMPLE

–

y = x2 –1

2

We could also graph y = |x2 – 1| using a different

1

+

x2 – 1

x2 – 1 is positive for x < –1 or x > 1 and zero for x = –1 or x = 1. For these values of x, |x2 – 1| = x2 – 1. x2 – 1 is negative for –1 < x < 1. For these values of x, |x2 – 1| = –|x2 – 1| = –x2 + 1.

Solution 2

–1

–1

O

1

x

–1

O

x

1

Draw the graph of f(x) = |x – 1| + |x + 2|. First we construct a sign table. x x–1 x+2

–2 – –

1 – +

y

+ +

When x < –2, f(x) = –x + 1 – x – 2 = –2x – 1. When –2 ≤ x < 1, f(x) = –x + 1 + x + 2 = 3. When x ≥ 1, f(x) = x – 1 + x + 2 = 2x + 1. So f(x) can be defined piecewise as x < −2 ⎧ −2 x − 1 if ⎪ f ( x) = ⎨ 3 if −2 ≤ x < 1 ⎪ 2 x +1 if x ≥ 1. ⎩

y = f(x)

3

–2

y = –2x – 1

–1

x

O 1

y=3

y = 2x + 1

Now we can draw the graph, shown opposite. 30

Algebra 10

Check Yourself 7 Draw the graphs. 1. y = |2x – 3|

2. y = |x3|

3. y = |x2 + 2x – 3 |

4. y = |x – 5| + |x + 3|

5. y = |x – 5| ⋅ x + 2x – 1

6. y = |2x + 1| + x – 3

C. SIGN FUNCTION sign function

Definition

Let f:

→

be a function.

y

The sign function sgn f(x) is defined as

EXAMPLE

27

Solution

28

Solution

Write the function f(x) = sgn(3 – x) as a piecewise function. Let us construct the sign table of (3 – x): 3

3–x

+

–

sgn(3 – x)

1

–1

⎧ 1 if x < 3 ⎪ Hence, sgn(3 – x) = ⎨ 0 if x= 3 ⎪ −1 if x > 3. ⎩

Solve the equations. a. sgn (2x + 1) = –1

b. sgn (x2 + 5x) = 0

c. sgn(

2x ) =1 x +1

1 . 2 b. If sgn (x2 + 5x) = 0 then x2 + 5x = 0, so x = –5 or x = 0.

a. If sgn (2x + 1) = –1 then 2x + 1 < 0, so x < –

c. If sgn( Functions

–1

For example, if x = –5 then sgn x = –1, if x = 0, sgn x = 0 and if x = 7, sgn x = 1.

x

EXAMPLE

x

O

The sign function helps us to determine the sign of a function on different intervals. The sign function is also called the signum function.

y = sgn x

1

⎧ 1 if f ( x) > 0 ⎪ sgn f ( x) = ⎨ 0 if f ( x) = 0 ⎪ −1 if f ( x) < 0. ⎩

2x 2x ) =1 then > 0, so x ∈ x +1 x +1

– [ – 1, 0]. 31

EXAMPLE

29

Solution

f:

→

, f(x) = x2 – x – 6 is given. Draw the graph of sgn(f(x)).

Let us construct the sign table of x2 – x – 6: x

–2

x2 – x – 6

3

+

sgn(x2 – x – 6)

y

–

1

+

–1

y = sgn f(x)

1

1

–2

⎧ 1 if x < –2 or x > 3 ⎪ So sgn f ( x) = ⎨ 0 if x = –2 or x= 3 ⎪ −1 if – 2 < x < 3. ⎩

3

O –1

x

The graph of y = sgn f(x) is shown opposite.

Check Yourself 8 1. Solve the equations. a. sgn (x2 – 1) = 0

b. sgn (2 – x) = 1

d. sgn(

2. Draw the graph of each function. b. f(x) = sgn (x2 – 5x + 6)

a. f(x) = sgn (2 – x)

x ) =1 x −1

c. f(x) = x – sgn (x)

Answers 1. a. {–1, 1}

b. x < 2

c.

– [0, 1]

D. FLOOR FUNCTION For any real number x, the greatest integer that is less than or equal to x is called the floor of x, denoted by x .

Definition

floor function Let f:

→

be a function. Then the floor function f(x) is defined as f ( x) if ⎧⎪ f ( x) = ⎨ ⎪⎩the greatest integer which is less than f ( x) if

32

f ( x) ∈ f ( x) ∉ . Algebra 10

Look at some examples of the floor function: 2

2.5

5

The floor function f : → , f(x) = x is also called the greatest integer function.

3

§2.5¨ = 2

5.75 6

§5.75¨ = 5

0 0.123

1

§0.123¨ = 0

–4 –3.7

–3

§–3.7¨ = –4

–1

0

§–0.4¨ = –1

–0.4

Similarly, 6 = 6, –5 = –5, 0.85 = 0, –2.5 = –3, π = 3, e = 2, –0.6 = –1 and –5.003 = –6.

Now let us draw the graph y = x in the domain [–3, 3]. Since the value y = x changes for each integer value of x, the integer values of x are important points for the function y = x . We say that these values are the crucial points of x . To draw the graph we have to consider the value of y between each of these crucial points: y if if if if if if if

–3 ≤ x < –2, –2 ≤ x < –1, –1 ≤ x < 0, 0 ≤ x < 1, 1 ≤ x < 2, 2 ≤ x < 3, x = 3,

y y y y y y y

= = = = = = =

x x x x x x x

= = = = = = =

–3; –2; –1; 0; 1; 2; 3.

3 2 1 –3

–2

–1 O

1

2

3

x

–1 –2 –3

y = §x¨

Remark

1. If x = t then we can write t ≤ x < t + 1 (t ∈ ). 2. For a ∈

and x ∈

For example, if if if if Functions

x x x x

, x + a = x + a.

= 3 then 3 ≤ x < 4; = –2 then –2 ≤ x < –1; = a then a ≤ x < a + 1; + 5 = 4 then x + 5 = 4, x = –1 and so –1 ≤ x < 0. 33

EXAMPLE

30

Solution

Draw the graph of f: [–4, 4] →

, f ( x) = x ⋅

x . 2

x takes an integer value. 2 x x = t then we can write t ≤ < t +1, i.e. 2t ≤ x < 2t + 2. For any real number x, if 2 2

First of all we have to find the crucial points at which

So the points 2t (t ∈ ) are the crucial points and we need to evaluate the function on the corresponding intervals: x if –4 ≤ x < – 2 then –2 ≤ < –1 2 x if –2 ≤ x < 0 then –1 ≤ < 0 2 x if 0 ≤ x < 2 then 0 ≤ <1 2 x if 2 ≤ x < 4 then 1 ≤ < 2 2 x then =2 if x = 4 2 We can now draw the graph on

and so f( x)= x and so f( x)= x and so f( x)= x and so f( x)= x and so f( x)= x

x 2 x 2 x 2 x 2 x 2

= x ⋅ (–2)= –2 x; = x ⋅ (–1)= – x; = x ⋅ 0= 0; = x ⋅1= x; = 4 ⋅2 = 8 .

each interval.

y y = f(x)

8 6 4 2

–4

O

–2

2

x

4

Check Yourself 9 1. Find each value. a. –2.1

b. –e

c. 0.9

d. ñ2

2. Solve the equations. a. x – 2 = –3

b.

c.

2x + 1 = 5

3. Draw the graph of each function. a. f(x) = 2x for –1 ≤ x ≤ 1 Answers 2. a. x ∈ [–1, 0)

34

b. x ∈ [2,

5 ) 2

e. π 3x – 1 4

f.

–π

= –1

b. f(x) = –x for –2 ≤ x ≤ 2 c. x ∈ [–1,

1 ) 3 Algebra 10

EXERCISES

2 .1

A. Piecewise Function 1. Given

5. Solve the equations.

⎧ x – 1 if ⎪ f ( x) = ⎨ x if ⎪ ⎩ 4x if

x> 2

f (0)+ f (2) 0 < x ≤ 2, find . f (3) – f (1) x≤0

2. Draw the graph of each piecewise function. ⎧⎪ a. f ( x) = ⎨ ⎪⎩

C. Sign Function

1

if x >1

–2 if x ≤ 1

⎧⎪ 2 x + 4 if b. f ( x) = ⎨ if ⎪⎩ – x

x >1 x ≤1

a. sgn x = 1

b. sgn (x – 1) = –1

c. sgn (x2 – 1) = –1

d. sgn (x3 – x) = 0

6. Draw the graphs. a. y = sgn x2 b. y = sgn (x – 1) x +1 c. y = sgn ( ) x–2

⎧ x2 +1 if x > 0 c. f ( x) = ⎪⎨ 2 ⎪⎩ x – 1 if x ≤ 0

D. Floor Function B. Absolute Value Function 3. Write each absolute value function as a piecewise function. a. f(x) = |x + 3| b. f(x) = |x| + x

7. Solve the equations. a.

x = –2

b.

x+1 =3

c.

x – 1 = –4

d.

2 x +1 =1 3

e.

1 – 3x =1 4

c. f(x) = |x2 – x – 2| d. f(x) = |x – 2| + |x – 3| e. f(x) = |x + 1| + |x – 1| f. f(x) = |x + 4| ⋅ x + x2 – 2x

8. Draw the graph of each function on the given interval. a. f(x) = x + 1 , x ∈ [–2, 2) b. f(x) = 2 – x , x ∈ [–3, 1]

4. Draw the graphs. a. y = |–x|

b. y = |2 – 4x|

c. f(x) = 2x – 1 , x ∈ [–1, 1]

c. y = |x2 – 1|

d. y = |x2 – 4x – 5|

d. f(x) = x x , x ∈ [–2, 2]

e. y = x|x + 1| + 3 f. y = |5x + 4| + 2x – 1 Functions

e. f(x) = x + x , x ∈ [–2, 2] 35

9. Find the values of x for which each function is undefined. ⎧ ⎪⎪ x2 f ( x) = ⎨ ⎪ ⎪⎩ x2

1 – 16

if x > 0

1 if x ≤ 0 –x–2

10. f and g are two piecewise functions defined as ⎧⎪ x2 – 1 if f ( x) = ⎨ ⎪⎩3x +1 if ⎧⎪ x +1 g( x) = ⎨ ⎪⎩ x + 2

x>2 x≤2

and

if x > 0 if x ≤ 0.

Find an expression for the function f(g(x)).

11. Find the inverse of the piecewise function f:

→

⎧⎪ 4 x – 1 if x >1 , f (x) = ⎨ ⎪⎩ 2 x +1 if x ≤ 1.

12. Write each function as a piecewise function. a. f(x) = |1 – x| b. f(x) = |x2 – 3x| – 4 c. f(x) = |x3| – x

13. Write each function as a piecewise function. a. f(x) = x ⋅ |x| b. f(x) = x + |x – 1| c. f(x) = x2 – |x2 – 2x – 3| d. f(x) = |x| + |x – 1| e. f(x) = |x – 2| – |x – 3| f. f(x) = |x2 – 1| + |x| 36

Algebra 10

1

CHAPTER REVIEW TEST

1. What is the domain of f ( x) = | 2x − 1| − | 3x +1|? A) [–2, 0]

B) (–2, 0)

C) [0, ∞)

D) (–∞, –2]

6. Which one of the following is the graph of an odd function? A)

B)

y

E)

2

2 2

A) [–4, –2)

D) (–2, 2]

C) (–2, 2)

y

C)

y

D) 2

–2

2

x

–2

O

x

–2

–2

3. What is the range of f(x) = –x + 5 for x ∈ [–5, 5]?

D) [5, 10]

x

2

E) (–2, 0]

B) [0, 5]

2 –2

O

A) [–5, 5]

O

–2

2−x ? x+ 2

B) [0, 2] – {1}

–2

x

O

–2

2. What is the domain of f ( x) =

y

E)

C) [–5, 10]

y 2

E) [0, 10]

2

–2 O

x

–2

4. Find the inverse of f: [1, ∞) → [4, ∞), f(x) = x2 – 2x + 5. A)

x−4

D)

B)

C)

x – 4 +1

x − 2 +1

E)

x2 − 2 +1

x − 4 −1

7. Given that f ( x) = find a. A) –2

B) –3

–2 x + 1 is a constant function, ax + 3

C) –4

D) –6

E) –10

5. Which one of the following is an odd function? A) f ( x) =

x2 | x|

C) f ( x) =

x5 x −x 3

B) f ( x) = 5 − x + 5+ x

to f ( x) = x – D) f(x) = x3 + x

E) f ( x) = Chapter Review Test 1

8. Which one of the following functions is equivalent

x2 x −1

x –1 for x < 1? | x – 1|

A) g(x) = x + 1

B) g(x) = x – 1

C) g(x) = 2x + 1

D) g(x) = x – 2

E) g(x) = x + 2 37

9. Which one of the following functions is equivalent ⎧⎪ x − 5 if x ≥ 5 to f ( x) = ⎨ ? ⎪⎩− x +5 if x < 5 A) g(x) = |x| – 5 B) g(x) = |x| + 5

C) g(x) = |x – 5|

13. Find the inverse of f: A)

3

x −1

3

B)

→ , f(x) = x3 – 3x2 + 3x. x − 1+1

x +1

D)

E)

C)

3

x +1 − 1

x − 1+1

D) g(x) = |x + 5|

E) g(x) = |2x – 5|

14. Which one of the following is an even function?

10. What is the domain of f ( x) = 5 x2 − 4x − 12 ? A) (–6, 2)

B) (–2, 6) D)

+

C)

A) f(x) = x7 – x3

B) f(x) = 2x

C) f(x) = xsin x

D) f(x) = xcos x

E) f(x) =

– (–2, 6)

cos x x3

E)

15. Given that 11. What is the domain of f ( x) = A) [–2, 2] D)

2 x +1 4 − x2

B) [0, –2] – (–2, 2)

3x2 − ( a − 2)x + 3 1 − x2 is an even function, find the value of a2 ⋅ f(a). f:

?

C) (–2, 2)

A) 5

− {−1, 1} → , f ( x) =

B) 10

C) –15

D) –20

E) 25

E)

16. Given f: [–a, a] → , which of the following must 12. What is the range of f ( x) = − x2 +5 x + 36 for the largest domain of f ? A) [0,

9 ] 2

D) (0, 38

B) [0, 15 ) 2

11 ] 2

C) [0, E) [0,

17 ) 2

13 ] 2

be an odd function? A) f(x) ⋅ f(–x) C)

f ( x)+ f ( −x) 2 1 D) f ( x)+ f ( ) x

B)

f ( x) − f ( − x ) 2

1 E) f ( x) − f ( ) x Algebra 10

A. LIMIT OF A POLYNOMIAL FUNCTION Consider the polynomial function f(x) = 2x. We are asked to investigate what happens to the value of f(x) as x gets closer to 2. We could begin by choosing a value of x which is close to 2, for example 1.5. We can calculate f(1.5) = 2 ⋅ 1.5 = 3. Now we choose a value which is closer to 2, for example 1.75: f(1.75) = 3.5. Continuing like this, we can make a table of values of f(x) as x gets closer to 2. x

1.5

1.75

1.8

1.9

1.95

2x

3

3.5

3.6

3.8

3.9

2

2.05

2.1

2.2

2.25

2.5

4.1

4.2

4.4

4.5

5

Using this table, we can guess that as x gets closer to (i.e. approaches) 2, the value of f(x) approaches 4. We say that 4 is the limit of f(x) = 2x as x approaches 2, and write lim( 2 x) = 4. In this notation, the arrow symbol (→) x→2

means ‘approaches’. x → 2 means x approaches the number 2. Notice that for f(x) = 2x, lim f ( x) = lim 2 x = 4, which is the same as f(2). Similarly, we can x→ 2

x→ 2

calculate lim f ( x) = lim 2 x = 6 = f (3) and lim f ( x) = lim 2 x = 24 = f (12), etc. In other x→ 3

x→ 3

x →12

x →12

words, in each case lim f ( x) = f ( c). In fact, this result is true for any polynomial function. x→ c

limit of a polynomial function

Definition

The limit of a polynomial function f(x) as x approaches a point c is f(c): lim f ( x) = f ( c). x→ c

In other words, for f(x) = anxn + an–1xn–1 + ... + a0, lim f ( x) = f(c) = ancn + an–1cn–1 + ... + a0. x→ c

For example, let us calculate the limit of f(x) = 2x when x approaches 5: lim f ( x) = lim 2 x = 2 ⋅ 5 = 10. x→5

EXAMPLE

1

x→5

Calculate the limits. a. lim(4 x – 1) x→ 2

Solution

b. lim( x2 + 3x + 2) x →–1

c. lim(3t – t2 ) t →4

These are all polynomial functions, so we can use lim f ( x) = f ( c). x→ c

a. lim(4 x – 1) = 4 ⋅ 2 – 1= 7 x→ 2

b. lim( x2 + 3x + 2) = (–1) 2 + 3 ⋅(–1)+ 2 =1 – 3+ 2 = 0 x → –1

c. lim(3t – t 2 ) = 3 ⋅ 4 – 4 2 =12 – 16 = –4. t→ 4

40

Algebra 10

Check Yourself 1 1. Given f(x) = 4x – 1, complete the table to find the limit of f(x) as x approaches 3. x f(x)

2.5 9

2.75

2.8

2.9

2.95

3

3.05

2. Calculate the limit of each polynomial function. a. lim5 x b. lim x(2 – x) x→0

x→ 3

d. lim 2 x( x +1)

e.

x→ a

lim 3t2 (2t – 1)

t → –3

3.1

3.2

3.25 12

c.

lim( x2 – 3x – 15)

f.

lim7

3.5

x→6

x→ 3

Answers 2. a. 0

b. –3

c. 3

d. 2a2 + 2a

e. –189

f. 7

B. LIMITS ON A GRAPH We can also use the graph of a function to study its limit as x approaches a certain point. For example, let us draw the graph of the function f: → , f(x) = x + 1 and use it to find lim f ( x). x→ 3

The notation f: → means a function from (the domain) to (the range). The notation lim f ( x) x →3

means the limit of f(x) as x approaches 3.

y

4.1

y = f(x) = x + 1

4.05 4.03 5

4.01

4 3.99

3

3.97

2

3.95 3.9

1

2.9

O

1

2.95

2.97

2

2.99

3

x

4

5

3.01

3.03

3.05

3.1

We can approach the point x = 3 from two directions: the right (as the values of x get gradually smaller) and the left (as the values of x get gradually bigger). In both cases, the limit of f(x) = x + 1 as x approaches 3 is 4: lim( x + 1) = 3+1= 4. x→ 3

Limit of a Function

41

EXAMPLE

2

Solution

A point at which we need to check the right-hand and the left-hand limits of a function is called a crucial point of the function.

⎧ 4 if ⎪ ⎪ ⎪ Given the piecewise function f : → , f ( x) = ⎨ 2 if ⎪ ⎪ ⎪ 2 x – 1 if ⎩ Let us first draw the graph of the function. As 5 we can see, x = is a crucial point in the graph. 2 5 When we approach from the right-hand side 2 5 (i.e. when x is greater than ) we use the 2 function f(x) = 2x – 1 and get lim(2 x – 1) = 4. x→

5 2

5 2 5 x = , find lim f ( x). 5 2 x→ 2 5 x> 2 x<

y

y = f(x)

6 5 4

So we can say that f approaches 4 from the

3

right-hand side. When we approach x from the 5 left-hand side (i.e. when x is less than ) we 2 use the function f(x) = 4, which is constant. Its

2 1 O

limit is 4 when x approaches from the left-hand

1

2 5 3 2

x

4

side. As a result, f approaches 4 as x approaches 5 from both sides, i.e. lim f ( x) = 4. 5 2 x→ 2

EXAMPLE

3

Solution

Given the piecewise function f :

⎧⎪ – x – 1 if x ≠ 2 , find lim f ( x). → , f ( x) = ⎨ x→ 2 if x = 2 ⎪⎩ 3

x = 2 is a crucial point. Notice that f(2) = 3 but 3 is not the limit of f at 2. This is because the limit is the value which f(x) approaches as x approaches 2. And in the graph we can see that the limit of f(x) when x approaches 2 is –3. As x gets closer to a point c, although the limit exists and approaches a number, at the point c a function may have a different value, or may not even be defined. What happens at the given point is not important for the limit at this point.

42

y

Let us draw the graph of f(x). 3 y = f(x)

2 1 –1

1 O

2

x

–1 –2 –3

Algebra 10

Note If we get different results for a limit when we approach it from the right and from the left, we say that the limit does not exist at this point.

Check Yourself 2 Graph each function and evaluate the given limit. 1. f:

→ , f(x) = 3,lim f ( x) x →2

2. f:

→

, f(x) = –x – 2, lim f ( x)

→

⎧ x – 1 if x >1 ⎪ ⎪ , f ( x) = ⎨ 1 if x =1, lim f ( x) x →1 ⎪ ⎪⎩ x2 +1 if x < 1

x →3

2

3. f:

⎧⎪ x2 if x ≠ –2 , f ( x)= ⎨ , lim f ( x) x → –2 ⎪⎩1if x = –2

→

4. f:

Answers 1. 3

2. –5

3. 4

4. does not exist

C. DEFINITION OF LIMIT We have seen how to calculate the limit of a polynomial function, and we have used graphs to calculate the limits of some other functions. However, we still do not have a general formula for the limit of a function. For example, consider the limit of the function f:

– {2} →

, f ( x) =

x2 – 4 when x x–2

x2 – 4 . f(x) is not a polynomial function and the graph of f(x) will be x→ 2 x – 2 x2 – 4 difficult to draw. Also, f ( x) = is undefined at x = 2. x–2 However, remember that when we calculate a limit we are examining the value of a function

approaches 2, i.e. lim

as it approaches a point, not the value at the point itself. So by applying simple factorization methods we can get lim x→ 2

x2 – 4 ( x – 2)( x + 2) = lim x → 2 x–2 ( x – 2) = lim( x + 2) = 2+ 2 = 4.

y

2 f(x) = x – 4 x–2

4

x→ 2

x–2 =1 because x is very close x–2 x–2 to 2 but not equal to 2, so is not zero x–2 divided by zero and so it can be simplified. As we

We can say that

can see in the graph of the function f:

– {2} → , f ( x) =

x2 – 4 , the limit of f(x) as x–2

3 2 1 O

1

2

x

x approaches 2 is 4. Limit of a Function

43

1. Neighborhood of a Number Let x0 be a real number and let a be a positive real number less than 1. Now consider the real number x between x0 – a and x0 + a such that x0 – a < x < x0 + a. In other words x is an element of the interval (x0 – a, x0 + a). This interval is called the a-n neighborhood of x0. x0 – a

x0

x0 + a

a-neighborhood of x0 x is a real number between x0 – a and x0 + a

For example, let us take x0= 5 and a = 0.1. The 0.1-neighborhood of 5 is the interval (5 – 0.1, 5 + 0.1) = (4.9, 5.1). 4

4.9

5

6

5.1

0.1-neighborhood of 5

Check Yourself 3 1. Write the 0.01-neighborhood of 7 as an interval and show it on a number line. 2. Write the 0.2-neighborhood of

5 as an interval and show it on a number line. 2

2. ε-neighborhood of a Number Now let x be an element of the ε-neighborhood of x0: ε is a Greek letter, pronounced ‘epsilon’. ‘ε-neighborhood’ is read as ‘epsilon-neighborhood’.

x ∈ (x0 – ε, x0 + ε). x0 – e

x0

x0 + e

e-neighborhood of x0

We can write the ε-neighborhood of x0 as an inequality: x0 – ε < x < x0 + ε. When we subtract x0 from all parts of this inequality we get: –ε < x – x0 < ε. We can write this inequality as an absolute value: |x – x0| < ε. 44

Algebra 10

x ∈ (3.8, 4.2)

For example, consider

x ∈ (4 – 0.2, 4 + 0.2) (4 – 0.2) < x < (4 + 0.2) –0.2 < x – 4 < 0.2 |x – 4| < 0.2. This means that x is a real number in the 0.2-neighborhood of 4. 3.8

3

4

4.2

5

x is here

3. Limit of a Function Definition

limit of a function Let f: → be a function and let x0 and L be real numbers. Given any ε about L if there exists a δ about x0 such that for all x, |x – x0| < δ ⇒ |f(x) – L| < ε then the limit of f as x approaches the point x0 is L, i.e. lim f ( x) = L. x → x0

Let us look at each part of the statement in turn. |x – x0| < δ δ is a Greek letter, pronounced ‘delta’. ‘δ-neighborhood’ is read as ‘delta-neighborhood’.

–δ < x – x0 < δ x0 – δ < x< x0 + δ x ∈ (x0 – δ, x0 + δ), i.e. x is in the δ-neighborhood of x0. x0 – d

x0

x0 + d

Similarly, |f(x) – L| < ε –ε < f(x) – L < ε L – ε < f(x) < L + ε f(x) ∈ (L – ε, L + ε) i.e. f(x) is in the ε-neighborhood of L. Limit of a Function

L+e L L–e

45

In other words, lim f ( x) = L means that for any small number ε there is another small x → x0

number δ such that any point in the δ-neighborhood of x corresponds to a point in the ε-neighborhood of L. y

y = f(x)

L+e

Any point in this neighborhood...

L L–e ... maps to a point in this neighborhood.

O

EXAMPLE

4

Solution

x0 – d

x

x0 x0 + d

Show that lim(3x – 5) = 4 by using the definition of limit. x→ 3

We will use the definition of limit with x0 = 3, f(x) = 3x – 5 and L = 4. To satisfy the definition, we need to show that for any ε > 0 there exists a δ > 0 such that for all x, |x – 3| < δ ⇒ |(3x – 5) – 4| < ε. y y = f(x)

Let us rewrite the second part of this statement: ε ε ε ε . 3 All these inequalities are equivalent. Now we |(3x – 5) |3x 3 ⋅ |x |x

– – – –

4| 9| 3| 3|

< < < <

need to show that for any ε > 0, there exists a δ > 0 such that for all x, |x – 3| < δ ⇒ |(3x – 5) – 4| < ε.

4+e 4 4–e 3 2 1 O

1

2 3 4 3– e 3+e 3 3

x

ε ε . Of course δ = is 3 3 not the only value of δ that will satisfy the

This is true if we take δ =

definition. Any smaller positive δ will work as well. However, the definition of limit asks us to find just one δ that satisfies the statement, so this is enough. 46

Algebra 10

EXAMPLE

5

⎧⎪ 2 x – 1 if x ≠ 2 → , f ( x) = ⎨ is given. Show that lim f ( x) = 3 by using the definition of x→ 2 ⎪⎩ 1 if x = 2

f:

limit. Solution

To prove that lim f ( x) = 3 we need to show that for all ε > 0 there exists a δ > 0 such that x →2

|x – 2| < δ ⇒ |f(x) – 3| < ε. Let us begin with the second part of the expression (|f(x) – 3| < ε) and let us try to show that for all ε > 0 we can find a number δ. |f(x) – 3| < ε ⇔ |(2x – 1) – 3| < ε ⇔ |2x – 4| < ε ⇔ |x – 2| < ε which is similar to 2 the first part of the expression (|x – 2| < δ).

y

Now we have |x – 2| <

ε Indeed, since is a real number we can 2 ε ε choose δ = and so |x – 2| < . 2 2 Now we have found that there is a real number ε δ which is equal to , so for any ε > 0 we can 2 ε choose δ = for this function. 2

ε 2

y = f(x)

4 3+e 3 3–e 2 1 –1

O

1

2 3 2– e 2+e 2 2

x

4

Moreover, we can say that for all ε > 0 there ε exists a δ = > 0 and |x – 2| < δ ⇒ |f(x) – 3| < ε. 2 ε ε For x ∈ (2 – δ, 2 + δ) = (2 – , 2 + ), f(x) ∈ (3 – ε, 3 + ε). 2 2 So as ε approaches zero, the point x approaches 2 and the value of f(x) approaches 3. Thus, lim f ( x) = lim(2 x – 1) = 3. x→ 2

x→ 2

Check Yourself 4 Use the definition of limit to prove each statement and find the value of δ. 1. lim(7 x – 2) = 5 x →1

2. lim(4 x + 3) = –5 x →–2

Answers ε ε 1. 2. 7 4 Limit of a Function

47

D. ONE-SIDED LIMITS Notice that in limit notation, x → x0 and x → x0+ mean different things. x → –x0 and x → x0– also have different meanings.

As we have already seen, sometimes the limit of a function can have two different values: one value when x approaches x0 from the right, and another when x approaches x0 from the left. hand When this happens, we call the limit of f as x approaches x0 from the right the right-h limit of f at x0 and write it as lim+ f ( x) . x →x 0

hand limit of f at x0 and write We call the limit of f as x approaches x0 from the left the left-h it as lim– f ( x) . y x →x 0

For example, consider the function y = f(x) shown opposite. Let us find the left-hand and right-hand limits at the points 2, 5 and 6:

5 4

a. lim– f ( x) = 3 and lim+ f ( x) = 5

3

b. lim– f ( x) = 2 and lim+ f ( x) = 2

2

c. lim– f ( x) = 3 and lim+ f( x) does not exist,

1

x→2

x→2

x→5

x→5

x→6

x→6

since f is not defined for x > 6.

y = f(x)

O

1

2

3

4

5

In other words the left-hand and right-hand limits as x → 2 are different; as x → 5 the left- and right-hand limits are the same, and as x → 6 only the left-hand limit exists. Definition

one-ssided limits 1. The limit of the function f(x) as x approaches x0 from the right equals L if for any ε > 0 there exists a δ > 0 such that for all x,

y L+e

x0 < x < x0 + δ ⇒ |f(x) – L| < δ,

L

i.e. for all x in the interval (x0, x0 + δ),

L–e

f(x) ∈ (L – ε, L + ε). O

2. The limit of the function f(x) as x approaches x0 from the left equals L if for any ε > 0 there exists a δ > 0 such that for all x, x0 – δ < x < x0

⇒ |f(x) – L| < ε, i.e.

for all x in the interval (x0 – δ, x0), f(x) ∈ (L – ε, L + ε).

48

x

6

x0 x0 + d

x

y L+e L L–e O

x0 – d

x0

x

Algebra 10

EXAMPLE

6

Solution

5 5 Find the one-sided limits at all the integer values of the graph of f : [– , ] → 2 2 the figure.

shown in

y

The integer values are {–2, –1, 0, 1, 2}, so we need to inspect these points. a. lim+ f ( x) = 0,

lim f ( x) = 0

x→ 2–

x→2

b. lim+ f ( x) =1,

x→1

c. lim+ f ( x) = –1, x→0

e.

lim+ f ( x) = 0,

1

lim– f ( x) = –1

x→0

lim– f ( x) =1

x→ –1

–2

O

–1

x → –1

lim+ f ( x) = 0,

x→ –2

2

lim– f ( x) = 0

x→1

d.

y = f(x)

3

1

x

2

–1

lim– f ( x) = 0.

x→ –2

existence of a limit The limit of a function f(x) at a point x0 exists if and only if the right-hand and left-hand y limits at x0 exist and are equal.

Definition

In other words, a b

lim f ( x) = L ⇔ lim+ f ( x) = L and

x → x0

x → x0

y = f(x)

lim f ( x) = L.

x2

x → x0 –

The left-hand limit and right-hand limit of a function are also called the one-ssided limits of the function.

EXAMPLE

7

Solution

For example, in the figure the function f has a limit at point x2, but it has no limit at point x1 because the left-hand and right-hand limits at x1 are different. f:

– {2} →

x1

c

if x > 2 ⎧⎪ x – 1 is given. Find lim f ( x). , f ( x) = ⎨ x →2 ⎪⎩ – x + 3 if x < 2 y

As we can see from the graph, the point x0 = 2 is the crucial point of f(x). Therefore, let us examine the one-sided limits at this point. lim+ f ( x) = lim( x – 1) = 2 – 1=1 + x→2

x→2

y = f(x) 3 2

lim f ( x) = lim(– x + 3) = –2+ 3 =1 –

1

Since lim+ f ( x) = lim– f ( x) =1, they exist

O

x→2 –

x

O

x→2

x→ 2

x→2

1

2

3

x

and are equal, and so lim f ( x) =1. x→ 2

Limit of a Function

49

Check Yourself 5 1. The graph of f: [1, 7] → , f(x) is shown in the figure. Find the one-sided limits at all integer values of the domain and find at which point f has a limit. ⎧⎪ x2 +1 if x >1 2. Given the function f: – {1} → , f ( x) = ⎨ , find x – 1 if x < 1 ⎪ ⎩ the limit of f(x) at the point x0 = 1.

y y = f(x)

7 6 5 4 3 2 1 O

1 2 3 4 5 6 7

x

⎧ x if –1 ≤ x < 0 or 0 < x ≤ 1 ⎪⎪ 3. The function f: → , f ( x) = ⎨ 1 if x = 0 is given. ⎪ ⎪⎩ 0 if x < –1 or x > 1 a. Find the one-sided limits at the points –1, 0, and 1.

b. At which point does the limit of f(x) exist? Answers 2. does not exist

E. LIMITS OF SPECIAL FUNCTIONS We have now learnt the definition and basic concepts of the limit of a function, and studied one-sided limits. In this section we will look at the limit of some special functions: the absolute value function, the sign function and the floor function. We know that at a given point, the limit of a function exists if the right-hand limit and the left-hand limit exist and are equal. We can evaluate a limit of an absolute value, sign or floor function at a point by treating the function as a piecewise function and checking the one-sided limits at the point. If the two limits exist and are equal, then we can say that a limit exists at the given point. EXAMPLE

8

Solution

f:

– {2} →

, f ( x) =

| x – 2| + x + 3 is given. Find lim+ f ( x) and lim– f ( x). x→2 x→ 2 x–2

Since the function f involves the absolute value expression |x – 2|, x0 = 2 is a crucial point for f. Let us begin by writing the function as a piecewise function. If x > 2, x – 2 > 0 and so |x – 2| = x – 2. Therefore f ( x) =

50

| x – 2| x– 2 + x+ 3 = + x + 3 =1+ x + 3 = x + 4. x–2 x–2 Algebra 10

Similarly, if x < 2, x – 2 < 0 and |x – 2| = –(x – 2) and so f ( x) =

y

| x – 2| – ( x – 2) + x+ 3 = + x+ 3 x–2 x–2

y=x+4

6 5 4 3 2 1

= –1+ x + 3 = x + 2. ⎧⎪ x + 4 if x > 2 In conclusion, f( x) = ⎨ ⎪⎩ x + 2 if x < 2.

–3 –2 –1 O

x

1 2 3

y=x+2

So the limits are lim f ( x) = lim( x + 4) = 2+ 4 = 6 +

x→2+

x→2

lim f ( x) = lim( x + 2) = 2+ 2 = 4. –

x→2 –

x→2

Remark

When we evaluate the right-hand and left-hand limits of special functions it is sometimes useful to define a positive real number h and look at what happens as h approaches zero. For h ∈

+

and h → 0, the following statements are true:

lim f ( x) = lim f ( a + h) and

x→ a+

h→0

x→ a

h→0

For example, when h ∈

+

EXAMPLE

9

Solution 1

– (x ® 3 )

x

(h ® 0)

3–h

h→0

lim f ( x) = lim f (3 – h).

x→3–

(x ® 3+)

3 + h (h ® 0)

and h → 0,

lim+ f ( x) = lim f (3+ h) and

x→3

x

3

lim– f ( x) = lim f ( a – h).

3

h→0

Given f(x) = sgn(4 – x), find lim+ f ( x) and lim– f ( x) and decide whether lim f ( x) exists or x→ 4 x→4 x→ 4 not. x0 = 4 is a crucial point.

y

When x > 4 (x → 4+), 4 – x is negative and so sgn(4 – x) = –1. So

1

lim+ f ( x) = lim+ sgn(4 – x) = –1.

x→4

O –1

x→4

y = sgn(4 – x) 1 2 3 4

x

When x < 4 (x → 4–), 4 – x is positive and so sgn(4 – x) = 1. So lim f ( x) = lim– sgn(4 – x) =1.

x→4–

x→4

Since the left-hand and right-hand limits are not equal, lim f ( x) does not exist. x →4

Limit of a Function

51

Solution 2

Let h > 0 be a very small real number. Instead of writing x → 4+, we can write x = 4 + h and consider h → 0: lim sgn(4 – x) = lim sgn(4 – (4+ h)) = lim sgn(– h).

x→4+

h→0

h→0

Since h is a very small positive number, –h is negative and so sgn(–h) = –1, i.e. lim sgn(– h) = –1. This is the right-hand limit. h→ 0

Similarly, lim– sgn(4 – x) = lim sgn(4 – (4 – h) = lim sgn( h) =1. h→0

x→4

h→0

Since h is a positive number, sgn(h) = 1. This is the left-hand limit. As before, lim+ f ( x) ≠ lim– f ( x) and so lim f ( x) does not exist. x→4

EXAMPLE

10

Solution

The function f:

x→4

x→4

→

, f(x) = 5x + 4 is given. Find lim f ( x). x→ 2

To find the limit as x approaches 2, we have to check the right-hand and left-hand limits: lim f ( x) = lim+ 5 x + 4 = lim 5(2+ h)+ 4 = lim 14+5 h =14

x→2+

h→0

x→2

h→0

lim f ( x) = lim– 5 x + 4 = lim 5(2 – h)+ 4 = lim 14 – 5 h =13.

x→2 –

h→0

x→2

h→0

Notice that since h is a very small positive number, 14 – 5h<14 and so 14 – 5h = 13. We can see that lim+ f ( x) ≠ lim– f ( x). x→2

x→2

So lim f ( x) does not exist. x→ 2

EXAMPLE

11

Solution

Find lim– x→1

| x – 1| . |1 – x|

x → 1– means x is less than 1. Therefore, x – 1 < 0 so |x – 1| = –(x – 1) = 1 – x, and 1 – x > 0 so |1 – x| = 1 – x. | x – 1| 1– x =1. = lim– = lim1 So lim– x→1 |1 – x| x→1 1 – x x→1–

52

Algebra 10

EXAMPLE

12

Solution

Find lim– sgn( x→1

x2 – 1 ). x+ 2

x

x2 – 1 . x+ 2

Let us make a sign table of

x2 – 1 x+ 2

x → 1– means that x approaches 1 from the left-hand side, and in this interval

–2 –

–1 +

1 –

+

x2 – 1 is x+ 2

negative. 2 So lim sgn( x – 1) = –1. x→1– x+ 2

EXAMPLE

13

Solution

Find lim– 2 x+1 . x→ 2

Let h > 0 be a very small real number. Then lim– 2 x +1 = lim 2(2 – h)+1 = lim 5 – 2 h = 4. h→ 0

x→2

h→ 0

Check Yourself 6 Evaluate the limits. 1. lim( x2 | x +1|)

2.

x →–3

lim

x +| x | x

2 3. lim | x – 4| – x→2 x–2

lim

|2 – x| sgn( x – 2)

6. lim+ sgn(3 – x) – x

lim+

1– x sgn( x2 – 1)

9. lim+ 2 x +1

x →–1

| x2 – 3x + 2| x –1

5.

7. lim+

| x |+2 sgn( x2 – 2)

8.

10. lim–

5x +1 16

11. lim( x + x + 2)

12. lim 8 x +1 x→4+ 3

14. lim– 2 x

15. lim x→ 4

4. lim– x→1

x→0

x→3

x→2+

x→1

x→3

x→2

x→ 2

2007 x – 1) 13. lim( – x→0

x→3

3x + 4 – x

Answers 1. 18 12. 11 Limit of a Function

2. 0

3. –4

13. –2

4. –1

14. 4

5. 0

6. –4

7. –2

8. 0

9. 5

10. 0

11. does not exist

15. does not exist 53

F. LIMITS INVOLVING INFINITY In this section, we will use the concept of infinity. Infinity is not a real number, but we can use it to describe a graph which continues without end in a positive or negative direction. For example, consider the function f : – {0} → – {c} graphed in the figure.

y

f(x) tends to positive infinity

How can we evaluate lim– f ( x) and lim+ f ( x)? x→0

x→0

As we can see in the graph, as we approach zero from the left-hand side the value of f(x) gets larger and larger. In other words, for any chosen number M, we can always find an x closer to zero on the left such that f(x) > M. We say that f(x) tends to (i.e. moves in the direction of) positive infinity, and write

c x ® 0–

x ® 0+

y = f(x) x

O

f(x) tends to negative infinity

lim f ( x) =+∞.

x→0–

Similarly, when we approach zero from the right-hand side the value of f(x) gets smaller and smaller. In other words, for any chosen number –M, we can always find an x closer to zero on the right such that f(x) < –M. We say that f(x) tends to negative infinity: lim f ( x) = – ∞.

x→0+

In both cases, f(x) has an infinite limit as x approaches a real number. Now, what about

y

lim f ( x) and lim f ( x)?

x →+∞

x→ – ∞

As we can see in the graph, as the value of x increases, the value of f(x) approaches the number c. In other words, for any chosen ε > 0, we can find a number M such that for all x > M, the value of f(x) will be in the ε-neighborhood of c, i.e. f(x) is getting closer and closer to c. So we can write

f(x) tends to c c x ® –¥

x ® +¥ x

O y = f(x)

lim f ( x) = c.

x→+∞

Similarly, as x gets smaller and smaller, the value of f(x) also approaches the number c as shown in the figure. For chosen any ε > 0, we can find a number N such that for all x < N, the value of f(x) will be in the ε-neighborhood of c, i.e. f(x) approaches the number c while x approaches negative infinity. So we can write lim f ( x) = c.

x→ – ∞

In both cases, as x approaches negative or positive infinity, f(x) approaches a real number. 54

Algebra 10

EXAMPLE

14

f:

– {0} →

a. lim+ f ( x) x→0

Solution

1 is given. Find the limits. x b. lim– f ( x) c. lim f ( x)

, f ( x) =

x →+∞

x→0

d. lim f ( x) x→ – ∞

1 First, let us graph f ( x) = to study the limits of the x function.

y y = f(x) = 1 x

a. As x approaches zero from the right-hand side, f(x) approaches +∞: lim

x→0+

1 =+∞. x

x ® 0– x ® 0+ x

O

For example, when we choose a positive number which is very close to zero such as x = 0.0001, we 1 get a large positive number: =10000. x

b. As x approaches zero from the left-hand side, f(x) approaches –∞: 1 lim = – ∞. x→0 – x For example, when we choose a negative number which is very close to zero such as x =–0.00001, we get a large negative number:

1 = –100000. x

c. As x approaches +∞, f(x) approaches zero: lim

x→+∞

1 = 0. x

For example, when we choose a large positive value of x such as x = 1000000, we get a small positive number close to zero: 1 = 0.000001. x

Limit of a Function

y y = f(x) = 1 x

x ® –¥ O

x x ® +¥

55

d. As x approaches –∞, f(x) approaches zero: 1 = 0. x→ – ∞ x lim

For example, when we choose a large negative value of x such as x = – 10000000, we get a small negative number close to zero: 1 = –0.0000001. x

As a consequence of results a and b, notice that lim x →0

EXAMPLE

15

Solution

Find the limit of the function f:

– {0} →

1 does not exist. x

, f ( x) =

2 x2 – x as x approaches +∞. x2 + 4

1 1 x2 (2 – ) (2 – ) 2 x2 – x x x = lim lim f ( x) = lim 2 = lim x→+∞ x→+∞ x + 4 x→+∞ 2 x→+∞ 4 4 x (1+ 2 ) (1+ 2 ) x x

0

1 (2 – ) 1 4 x = 2. → 0 and 2 → 0, so lim As x → +∞ we have x→+∞ 4 x x (1+ 2 ) x 0

EXAMPLE

16

Solution

Find the limit of the function f ( x) =

1 as x → 1. x –1 y

The function f is not defined at x0 = 1 and so x0 = 1 is a crucial point. So let us check the right-hand and left-hand limits. Let h > 0 be a very small number, then lim f ( x) = lim+

x→1+

x→1

y= 1 x–1

1 1 1 = lim = lim =+ ∞, x – 1 h→0 (1+ h) – 1 h→0 h

1 O

lim f ( x) = lim–

x→1–

x→1

1 1 1 = lim = lim = – ∞. h → 0 h → 0 x–1 (1 – h) – 1 –h

–

x®1

x ® 1+

x

The right-hand and left-hand limits are not equal, so lim f ( x) does not exist. x →1

56

Algebra 10

Remark

For some functions f:

→

, the following infinite limits are possible:

a. lim f ( x) =+∞

b. lim f ( x) = – ∞

c. lim f ( x) =+∞

d. lim f ( x) = – ∞.

x →+∞

x →+∞

x→ – ∞

x→ – ∞

Look at the examples. y

y

y

y = –x2

x ® –¥

x ® –¥ O

x

x

O

x ® +¥

x ® –¥

x ® +¥

y = x3

x ® +¥ x

O

y = x2

Check Yourself 7

y

1. The graph of a function f(x) is shown opposite. Find each limit. a.

lim f ( x)

x →+∞

c. lim f ( x) x →1

2. Calculate the limits. 2x + 3 a. lim x →+∞ 5 x + 4 c.

x2 – 2 x + 3 x →+∞ 2 x2 + 3x – 1 lim

b. lim f ( x)

2

x→ – ∞

d. lim f ( x) x →3

O

b.

Limit of a Function

lim( x5 + x4 + x3 )

x

3

x3 +1 x →+∞ x4 | x| x → – ∞ | x |+1

c. lim+ x→ 2

1 x –4 2

e.

d. lim x→ e

4. Calculate the limits. x→ – ∞

1

lim

d. lim

3. Calculate the limits. 1 1 a. lim 2 b. lim– x →0 x x→1 x – 1 a.

y = f(x)

b. lim(1 – x – x2 ) x →∞

c.

1 – x2 x → – ∞ 1+ 3x2

f.

lim

1 1 – ln x

e. lim x→ 2

lim

x→ – ∞

x2 – 2 4x – 2

5x – 1 2–x

3 lim(2 x – ) x

x→ – ∞

57

Answers 1 2 1 b. 0 c. d. 1 e. – 1 f. – 3. a. +∞ b. – ∞ 4 5 2 3 c. +∞ d. does not exist e. +∞ 4. a. – ∞ b. – ∞ c. – ∞

1. a. 2 b. 2 c. +∞ d. does not exist 2. a.

Theorem

limit combination theorem Let f(x) and g(x) be functions such that lim f ( x) = a and lim g( x) = b. Then x → x0

x → x0

a. lim[ f ( x)+ g( x)]= a + b x→ x0

b. lim[ f ( x) – g( x)]= a – b x→ x0

c. lim[ f ( x) ⋅ g( x)]= a ⋅ b x→ x0

d. lim[ x → x0

f ( x) a ]= ( b ≠ 0) g( x) b

e. lim k ⋅ f ( x) = k ⋅ a (k ∈ ). x → x0

EXAMPLE

17

Given f(x) = 5 and g( x) =

1 (x ≠ 0), evaluate the limits. x

a. lim[ f ( x) + g( x)]

b. lim[ f ( x) ⋅ g( x)]

x →–2

Solution

x → –2

x →+∞

x →–2

x →–2

– {0} →

x→0

x→+∞

x→+∞

x→+∞

f ( x) ] g( x)

1 1 9 = 5+( – ) = x 2 2

1 =5 ⋅ 0=0 x

f ( x) 5 ]= lim = lim 5 ⋅ x = 5 ⋅ lim x = 5(–3) = –15 x → –3 x→–3 1 x→–3 g( x) x

a. lim+ f ( x) 58

x → –2

lim [ f ( x) ⋅ g( x)]= lim f ( x) ⋅ lim g( x) = lim 5 ⋅ lim

x→ –3

f:

x → –2

x →+∞

c. lim[

18

x →–3

a. lim[ f ( x)+ g( x)]= lim f ( x)+ lim g( x) = lim 5+ lim b.

EXAMPLE

c. lim[

x→+∞

, f ( x) =

1 is given. Calculate the limits. 1+ 22 / x

b. lim– f ( x) x→0

c. lim f ( x) x→0

Algebra 10

Solution

a. lim+ f ( x) = lim+ x→0

x→0

lim 1 1 x→0+ = 2/ x 1+ 2 lim+ 1 + lim+ 2 2 / x x→0

+ As x → 0 ,

x→0

lim 1 2 1 x→0+ = = 0. → +∞ and so 2 2 / x →+ ∞. So lim+ 1 + lim+ 2 2 / x 1 + ∞ x x→0

f ( x) = lim– b. xlim →0 – x→0

lim 1 1 x→0 – = 2/ x 1+ 2 lim– 1+ lim– 2 2 / x x→0

– As x → 0 ,

x→0

x→0

lim 1 2 1 1 1 x→0– → – ∞ and so = = = =1. x lim– 1+ lim– 2 2 / x 1+ 2 −∞ 1+ 1 1+0 x→0 x→0 2∞

c. Since lim+ f ( x) ≠ lim– f ( x), lim f ( x) does not exist. x→0

x→0

x→0

Check Yourself 8 1. Given lim f ( x) = 2, find lim x→ a

x→ a

f 2 ( x) − f ( x) . f ( x)+ 3

2. f(x) = sin x + 2 and g(x) = cos 2x are given. Find limπ x→

2

f ( x) ⋅ g( x)+ f ( x) . ⎛g⎞ ⎜ ⎟ ( x) ⎝f⎠

3. Calculate each limit by using the limit combination theorem. 2 1+ x–2 x a. lim 2 b. lim c. lim(3 x +1)2 x→ – ∞ 4 x →1 x + 4 x →1 5+ x

d. lim(6 x3 + 4 x2 – 3) x →–1

Answers 1.

Limit of a Function

2 5

2. 0

3. a. –

1 5

b.

1 5

c. 16

d. –5

59

EXERCISES

2 .1

A. Limit of a Polynomial Function

15. Use the definition of the limit of a function to prove each statement.

In questions 1-7, calculate the limits.

1. lim(3 x + 2)

2. lim( x2 – 2 x)

3. lim 3(2t – 1)( t +1)

4. lim17

5. lim(2 k +1)

x +1) ⋅ x2 6. lim(2 x→ a

x→5

x →–4

t→2

x→ –

5=5 c. lim x →3

16.

y 2

x2 – 2 b +1) 7. lim( x→ b

B. Limits on a Graph

→

, f(x) = 4x – 7, lim f ( x)

→

⎧ x if x > 0 ⎪⎪ , f(x) = ⎨ 3 if x = 0 ⎪ 2 ⎪⎩ – x if x < 0

y = f(x)

1

–3 –2

In questions 8-10, draw the graph of each function and calculate the given limits.

8. f:

2

D. One-Sided Limits

x→ 4

x →–2

x + 3) = 2 b. lim(2 1

2 x = 10 a. lim x→5

–1

1

x

O

2

3

–1 –2

x→3

–3 2

9. f:

a. lim f ( x) x→0

10. f:

→

a. lim+ f ( x)

b. lim+ f ( x)

c. lim– f ( x)

d. lim+ f ( x)

e. lim– f ( x)

f.

x→–1

c. lim f ( x)

x→ 2

if ⎧ 2 ⎪ , f(x) = ⎨ 1 if ⎪ ⎩ x + 3 if

a. lim f ( x) x →–5

b. lim f ( x)

The figure shows the graph of a function f(x). Find each limit.

x →–2

x→2

x→0

x > –3

x→ –2

x→ –2

17.

x = –3

y y = f(x)

x < –3

b. lim f ( x) x →–3

2

c. lim f ( x) x →–1

C. Definition of Limit 11. Write the 0.3-neighborhood of 12 as an interval. 12. Write the 0.1-neighborhood of –5 as an interval and show it on a number line.

13. Write the 0.01-neighborhood of 4 as an interval. 14. Write the 0.05-neighborhood of 5 as an interval. 60

lim f ( x)

x→ –1–

1 –3

–2

–1

O

1

2

3

4

x

–1

The graph of a function f is shown in the figure. For which integer values of p in the interval 7 9 (– , ) does lim f ( x) exist? x→ p 2 2 Algebra 10

18. The function f:

→ , if x >1

x ⎧ ⎪ 2 if x =1 or x = –1 ⎪ f(x) = ⎨ 2 ⎪ – x +1 if –1 < x <1 ⎪ if x < –1 ⎩ –x is given.

21. f:

b. Find at which point(s) the limit does not exist.

→

x> 3 x= 3 1≤ x<3 x <1

is given. Calculate the limits.

a. Find the one-sided limits as x → –1, x → 0 and x → 1.

19. The function f:

→

⎧ x – 2 if ⎪ if ⎪ 3 , f(x) = ⎨ ⎪ – x + 4 if ⎪ if ⎩ 1

a. lim f ( x)

if x >1 ⎧ –3 ⎪ 2 if x =1 or x ≤ –2 f(x) = ⎨ ⎪ 2 ⎩ x – 4 if –2 < x <1

x→ 3

22. The figure shows the graph of a function g(x). Find the sum of the left-hand limits of g(x) at the integer values in the interval (0, 8).

,

b. lim f ( x)

x →1

is given.

y 8 7 6 5 4 3 2 1 O

y = g(x)

1 2 3 4 5 6 7 8

a. Find the limits as x → 0 and x → 1. b. Find at which point(s) the limit does not exist.

E. Limits of Special Functions 23. Calculate the limits.

20.

y 4 y = f(x) 3 2

a. lim | x – 2| x→2+ |2 – x|

b. lim|9 – x2 |

c. lim |2 x | x→0+ x + x

d. xlim → 5–

e. lim+ π x→ 2

|cos x| cos x

x→ 3

| x – 5| x–5

f. lim | x| x→0– x

1 –3

–2

–1

O

1

2

3

x

–1

24. Calculate the limits. x2 – sgn( x – 2)) a. lim( – x→2

–2

The graph of a function f: (–3, 3) → is shown in the figure. At which integer value(s) of the domain does a limit exist? Limit of a Function

x)) b. lim(sgn(cos + x→

π 2

c. lim+

|4 – x2 | sgn|2 – x| – x

d. lim –

e. lim

3x + 2 sgn(– x2 +6 x – 9)

f. lim

x→2

x→ 3

x→5

x →1

x2 – 2 x x – 3sgn( x – 5)

1– x sgn( x2 – 3 x + 2) 61

x

25. Calculate the limits. x + 3sgn x) a. lim( – x→2

c. lim+ x→π

sin x sin x +1

e. lim x2 − 3 x→ 3

28. Calculate the limits. b. lim( x + x)

a. lim+

1 x2 – 4

b. lim

1 | x|

cos x + cos| x|) d. lim( +

c. lim–

x x –9

d. lim

x3 + 2 x | x – 1| – 6

x→ 3

x→2

x→π

f. lim+ x→3

3– x x–3

x →0

2

x→3

x→ 3

e. lim( x – tan x) – x→

π 2

g. lim– x5 x→2

29. f:

, f ( x) =

→

x+ 2 is given. x( x +1)( x – 2)

Calculate the limits. a. lim f ( x)

b. lim– f ( x)

F. Limits Involving Infinity

c. lim+ f ( x)

d. lim– f ( x)

26. The graph of a function f

e. lim+ f ( x)

f. lim– f ( x)

g. lim+ f ( x)

h. lim f ( x)

x→ – ∞

x→ –1

y

x →+∞

1

x →+∞

c. lim f ( x)

x→ 2

x→2

a. lim f ( x) x→ – ∞

x→0

x→0

is shown in the figure. Evaluate the limits.

b. lim f ( x)

x→ –1

–2

–1 O –1

1

x y = f(x)

30. Calculate the limits. a. lim (– x2 +1) x →+∞

x →1

–2

d. lim f ( x)

2 x3 – x2 +1 x → – ∞ x2 + 2 x +1

c. lim

x →–1

e. lim f ( x)

x3 x → – ∞ x +5

b. lim

d. lim

x →+∞

2

1 1

1– 5x

x →–2

31. Evaluate lim x →2

f ( x)+ f ( x) given lim f ( x) = 4. x→ 2 f ( x) – 5

32. Use the limit combination theorem to calculate 27. Calculate the limits. 2 x2 +1 a. lim x →+∞ 3x2 + 4

c. lim( x→ – ∞

e. lim

x→ – ∞

62

2

x –1 ) x2 4 x2 + x – 5 3

x3 + 1

1 b. lim ( +1) x →+∞ x

d. lim( x→ – ∞

2| x | ) x

f. lim (1+ cos 1 ) x →+∞ x

each limit. 2x – 1 a. lim 2 x→ 3 x + 1 c. lim 5 x →π

e. lim x →7

3

x–π x+ π

b. lim(4 x2 – 3 x+6) x→ 2

d. lim 2 x2 + 4 x→ 2

x2 – 1 x –1 Algebra 10

So far we have studied the concept and formal definition of the limit of a function. We have seen that even if a given function f is not defined at x0, in some cases the limit of the function may exist as the point x approaches the point x0 or infinity. For example, the function f ( x) = value

x2 – 9 is not defined at x0 = 3: when x0 = 3, we get the x–3

0 . But we know that lim f ( x) is not the same as f(3). It doesn't matter what happens x→ 3 0

at x0 = 3; the important thing is what happens as x approaches this point. So the limit of a function may exist even at undefined points. Similarly, the function f ( x) =

x2 + x +1 ∞ approaches the value as x approaches infinity. 2 2x – 3 ∞

∞ 0 and are two examples of indeterminate forms. In the following section we will look at ∞ 0 0 ∞ how to evaluate limits which have the indeterminate forms , , 0 ⋅ ∞, ∞ – ∞, and 1∞. 0 ∞

A.

0 AS A LIMIT 0 Let f and g be two functions and let x0 ∈ \ such that f(x0) = 0 and g(x0) = 0. Then lim x → x0

f ( x) 0 has the indeterminate form . g( x) 0

In this case there exists a function h(x) which is a common factor of the functions f and g such that f(x) = f1(x) ⋅ h(x), g(x) = g1(x) ⋅ h(x) and h(x0) = 0. So we can write lim

x → x0

f ( x) ⋅ h( x) f ( x) = lim 1 . g( x) x → x0 g1( x) ⋅ h( x)

Since x ≠ x0 (it is very close to, but not equal to, x0) we can cancel the factors h(x) and so lim

x → x0

f1( x) ⋅ h( x) f ( x) f1( x0 ) = lim 1 = g1( x) ⋅ h( x) x→ x0 g1( x) g1( x0 )

which is the result of the limit. Limit of a Function

63

EXAMPLE

19

Solution

Find lim x→ 3

x2 − 9 . x−3

0 As x approaches 3, the quotient approaches the indeterminate form . However, by 0 factoring the expression in the numerator we get lim x→ 3

x2 − 9 ( x − 3)( x + 3) = lim = lim( x+ 3) = 6. x→ 3 x→ 3 x−3 ( x − 3)

Remember that we can cancel the factor

( x – 3) because x ≠ 3 at the limit (it is only ( x – 3)

approaching this point).

EXAMPLE

20

Solution

Find lim x→ 2

x2 − 5x +6 . x2 − x − 2

0 as x approaches 2, so we need to factor the numerator 0 and denominator and find the common factor:

This limit has the indeterminate form

lim x→ 2

EXAMPLE

21

Solution

Find lim x →1

x2 – 5x +6 ( x – 2)( x – 3) ( x – 3) 1 = lim = lim =– . x → 2 ( x – 2)( x +1) x → 2 ( x +1) 3 x2 – x – 2

x −1 . x3 − 1

0 as x approaches 1. 0 We can begin by multiplying both the numerator and the denominator by the conjugate of ñx – 1:

The limit has the indeterminate form

lim x →1

x –1 ( x – 1)( x +1) ( x − 1) = lim 3 = lim 3 . 3 x → x → 1 1 x –1 ( x – 1)( x +1) ( x − 1)( x +1)

We know that x3 – 1 = (x – 1)(x2 + x + 1), so lim x →1

( x – 1) 3

( x – 1)( x +1)

= lim x →1

= lim x →1

64

( x − 1) 2

( x – 1)( x + x+1)( x +1) 1

1 = . ( x + x +1)( x +1) 6 2

Algebra 10

Check Yourself 9 Calculate the limits. 1. lim

x2 – 2 x – 8 x2 – 3 x – 4

4. lim

x+ 2 – 2 x–2

x→ 4

x→ 2

Answers 6 2 1. 2. 5 9

3.

2. lim x→ 3

x2 – 9 x3 – 27

5. xlim →16 4 2 3

4.

1 4

5. 4

3. lim x →1

4 – 2x x→ 2 4 x – 16

x–4

6. lim

x–2

6. –

1. The Indeterminate Form

x3 – x2 – x +1 3 x2 – 6 x + 3

1 8

0 in Trigonometric Functions 0

As x approaches a point x0, the limit of a trigonometric function is the image of x0. π For example, lim cos x = cos0 =1, lim sin x = sin0 = 0, and limπtan x = tan =1. x→0 x→0 4 x→ 4

But what about a limit such as lim x →0

sin x ? x

When we are calculating the limit of a trigonometric function which involves the indeterminate form

sin x 0 =1. , we can use the rule lim x →0 x 0

Conclusion 1. lim

tan x =1 x

2. lim

sin ax a tan ax a = and lim = x → 0 bx b bx b

3. lim

sin ax a tan ax a = and lim = x → 0 sin bx b tan bx b

4. lim

sin ax a tan ax a = and lim = . x → 0 sin bx b tan bx b

1. lim

tan x sin x 1 sin x 1 ⋅ ⋅ lim = lim = lim =1 ⋅ 1=1. x→0 x x cos x x→0 x x→0 cos x

x→0

x→0

Proof

x→0

2. As x → 0, let ax = u so x = lim x→0

Limit of a Function

x→0

x→0

u and consider u → 0. a

sin ax 1 sin ax 1 sin u a sin u a a = ⋅ lim = ⋅ lim = ⋅ lim = ⋅1= bx b x→0 x b u →0 u b u →0 u b b a 65

tan ax tan ax ⋅ ax tan ax a 1 a a = lim ax = lim ax ⋅ = ⋅ = 3. lim x → 0 tan bx x → 0 tan bx x → 0 tan bx b 1 b b ⋅ bx bx bx sin ax sin ax sin ax ax . lim sin ax a a x→0 ax a 1 a ax ax = lim = ⋅ lim = ⋅ = ⋅ = 4. lim x→0 tan bx x→0 tan bx b x→0 tan bx b tan bx b 1 b bx ⋅ lim x→0 bx bx bx

EXAMPLE

22

Solution

EXAMPLE

23

Solution

Find lim x→ 2

4 – x2 (2 – x)(2+ x) (2 – x) ⋅ lim(2+ x) =1 ⋅ 4 = 4 = lim = lim 2 x → 2 sin(2 – x) x→ 2 x → sin(2 – x) sin(2 – x) x→ 2

lim

Find lim x→π

x→π

24

tan x . 3( x – π)

Since tan x = –tan(π – x), lim

EXAMPLE

4 – x2 . sin(2 – x)

tan x – tan( π – x) 1 tan( π – x) 1 1 = lim = lim = ⋅1= . 3( x – π) x→π –3( π – x) 3 x→π ( π – x) 3 3

Find lim x→0

sin 2 3x2

x 2. 2

Solution

EXAMPLE

25

Solution

x x x x⎞ ⎛ sin 2 sin 2 sin ⎟ ⎜ 1 1 1 1 2 = lim 2 = ⋅ lim 2= 2 = 1 ⋅12 = 1 lim lim ⎜ ⎟ 2 2 x → 0 3 x2 x → 0 x → 0 x → 0 x 3 4x 3 4 12 12 12 ⎛ x⎞ ⎜⎜ ⎟⎟ ⎜ ⎟ 2 ⎝ ⎠ 4 ⎝2⎠ sin 2

Evaluate lim x→0

1 – cos 2 x . 8 x2

Since cos 2x = 1 – 2sin2 x, 1 – cos 2 x 1 – (1 – 2 sin 2 x) 1 – 1+ 2 sin 2 x sin 2 x 1 sin x 2 = lim( = lim = lim = lim ) 2 2 2 2 x→0 x → x → x → x → 0 0 0 0 8x 8x 8x 4x 4 x

lim

= 66

1 2 ⋅ 1 = 4. 4 Algebra 10

EXAMPLE

26

Evaluate lim cos x . π x→ π – 2 x 2

Solution

Since cos x = sin(

π − x) we can write 2

π 1 sin( – x) sin ( π – 2 x) cos x 1 2 2 limπ = limπ = limπ = . ( π – 2 x) ( π – 2 x) 2 x→ ( π – 2 x) x→ x→ 2

2

2

Check Yourself 10 Calculate the limits. tan 2 x 1. lim x → 0 tan5 x

sin 2 2 x x→0 x2

sin(2 x − 2) x →1 4x − 4

sin5 x 5x ⋅ cos5 x

4. lim

5. lim

6. lim x→0

tan πx 7. lim x →1 1 − x

3x2 8. lim x → 0 1 − cos 2 x

3π ) 2 9. lim x → 0 sin( π − 2 x)

Answers 1 2 1. 2. 3 5

B.

x 2 3. lim x → 0 3x sin

sin x 2. lim x → 0 tan 3 x

3.

1 6

4.

1 2

5. 4

6. 1

7. –π

cos(3 x +

8.

3 2

9.

3 2

∞ AS A LIMIT ∞ Let f and g be two functions and let x0 ∈ \ such that as x approaches x0, both f(x) and g(x) approach positive or negative infinity, i.e. lim f ( x) = ±∞ and lim g( x) = ±∞. x→ x0

Then lim x→ x0

x→ x0

∞ f ( x) has the indeterminate form . We can find the limit of such functions as g( x) ∞

follows:

Limit of a Function

67

Let P(x) = anxn + an–1xn–1 + ... + a0 and Q(x) = bmxm + bm–1xm–1 + ... + b0 be two polynomial functions, then a xn + an −1x n −1 +...+ ao P( x) = lim n m m −1 x →±∞ Q( x) x→±∞ b x + b +...+ bo m m −1x lim

0

0

a a xn ( an + n −1 +...+ on ) n x x = lim an x . = lim x →±∞ m x→±∞ b x m b b m x ( bm + m −1 +...+ mo ) x x 0

0

As x approaches ±∞, the limit of each rational expression approaches zero, so we have eliminated the rational expressions. In conclusion, we can write ⎧ an if n = m ⎪b n m ⎪ a x P( x) lim = lim n m = ⎨ x →±∞ Q( x ) x→±∞ b x if n < m m ⎪0 ⎪ ⎩ ±∞ if n > m. The degree of a polynomial is the highest power of the variable in the polynomial. The leading coefficient is the coefficient of the term with the highest degree. For example, x – 3x2 has degree 2 and its leading coefficient is –3.

EXAMPLE

27

Solution

This is the same as saying the following: 1. If the degrees of the polynomials are equal then the limit is the ratio of the leading coefficients. 2. If the degree of the polynomial in the denominator is greater than the degree of the polynomial in the numerator then the limit is zero. 3. If the degree of the polynomial in the denominator is less than the degree of the polynomial in the numerator then the limit is infinite (i.e. it approaches positive or negative infinity). Evaluate the limits. 5 x 3 – 2 x2 + x – 1 x →∞ 4x3 – x2 +12

–2 x3 – 1 x →∞ x5 + x3 +1

a. lim

b. lim

The indeterminate form is

∞ . ∞

x2 + 2 x →−∞ x + 4

c. lim

5 x 3 − 2 x2 + x − 1 5 = because the polynomials have the same degree. Alternatively, we x →∞ 4x3 − x2 +12 4

a. lim

can calculate 5 x 3 − 2 x2 + x − 1 lim = lim x →∞ x→∞ 4x3 − x2 +12 68

2 1 1 + − ) 3 x x2 x3 = lim 5 x = 5 . 3 x→∞ 4 x 1 12 4 x3 (4 − + 3 ) x x

x3 (5 −

Algebra 10

b. lim x→∞

−2 x3 − 1 = 0 because the degree of the polynomial in the denominator is bigger. We x5 + x3 +1

can also calculate this limit: 1 x3 ( −2 − 3 ) −2 x3 − 1 −2 x3 −2 x lim 5 = lim = lim 5 = lim 2 = 0. 3 x →∞ x + x +1 x →∞ 5 x →∞ x x →∞ x 1 1 x (1+ 2 + 5 ) x x x2 + 2 = −∞ because the degree of the polynomial in the numerator is bigger. We can x→−∞ x + 4

c. lim

also calculate this directly: x2 + 2 lim = lim x →−∞ x + 4 x→−∞

EXAMPLE

28

Calculate the limits. a. lim x →∞

Solution

2 ⎞ ⎛ x2 ⎜ 1+ 2 ⎟ x2 x ⎠ ⎝ = lim = lim x = −∞. x→−∞ x x→−∞ 4⎞ ⎛ x ⎜ 1+ ⎟ x⎠ ⎝

2x + 4

b. lim

9 x2 + 2

x →−∞

x − 4x2 − 1 x2 + 2 x + 2

a. The indeterminate form is

∞ . ∞

4⎞ ⎛ x ⎜ 2+ ⎟ x⎠ ⎝ lim = lim 2 x →∞ x→∞ 2 9x + 2 ⎛ 9x2 ⎜ 1+ 2 9 x ⎝ 2x + 4

b. The indeterminate form is

4⎞ 4⎞ ⎛ ⎛ x ⎜ 2+ ⎟ x ⎜2+ ⎟ 2 x⎠ x⎠ ⎝ ⎝ = lim = lim = x→∞ x→∞ 3 2 2 ⎞ 3 x 1+ 2 3 x 1+ 2 ⎟ 9 9 x x ⎠

∞ . ∞

1 ⎞ 1 ⎞ ⎛ ⎛ x − x ⎜4 − 2 ⎟ x − ( −x) ⎜ 4 − 2 ⎟ x ⎠ x ⎠ x − 4x − 1 ⎝ ⎝ lim = lim = lim 2 x →−∞ x→−∞ x→−∞ 2 2 2 2 x + 2x + 2 x 1+ + 2 − x 1+ + 2 x x x x 2

⎛ 1 ⎞ 1 x ⎜⎜ 1+ 4 − 2 ⎟⎟ x ⎠ 1+ 4 x2 = lim ⎝ = lim = = −3 x →−∞ x→−∞ − 1 2 2 2 2 − x 1+ + 2 −x 1+ + 2 x x x x x+ x 4 −

Limit of a Function

69

EXAMPLE

29

Solution

2 x +5 x+1 . x →∞ 3x +5 x

Find lim

The indeterminate form is

∞ . ∞

2 5x ⎛⎜ ( )x +5 ⎞⎟ 2 x +5 x+1 ⎝ ⎠ = 5, since lim( 2 ) x = 0 and lim( 3 ) x = 0 . 5 lim x = lim x→∞ 3 +5 x x→∞ x→∞ 5 x→∞ 5 3 ⎛ ⎞ 5x ⎜ ( )x +1⎟ ⎝ 5 ⎠

Check Yourself 11 Calculate the limits. 1 – x2 +7 x3 1. lim x →∞ 5+ x – x3 4. lim

x5 + 2 x +7 x →−∞ x2 +12 x – 1

3x2 +1

x →−∞

5. lim

2

x →∞

4x + 4 x tan 2 7. lim x→π cot( x − π)

2. –∞

3. lim

x − 9 x2 − 1

6. lim x→0

2

x +1+ 3x

x2 + cos x x →∞ 3 x2

3. 0

4. ∞

5. –

1 2

6.

6 5

cot 5 x cot 6 x

7 x+2 + 3x x→∞ 5 x + 7 x

8. lim

Answers 1. –7

x2 + 2 x − 12 x →−∞ 1 − x3

2. lim

9. lim

7. –2

8.

1 3

9. 49

C. 0 ⋅ ∞ AS A LIMIT Let f and g be two functions and let x0 ∈ \, and let us assume that as x approaches x0, f(x) approaches zero but g(x) approaches positive or negative infinity, i.e. lim f ( x) = 0 and lim g( x) = ±∞.

x → x0

x → x0

Then lim[ f ( x) ⋅ g( x) has the indeterminate form 0 ⋅ ∞. x→ x0

We can find the limit of such functions by transforming the indeterminate form 0 ⋅ ∞ into the 0 ∞ indeterminate form or as follows: 0 ∞ f ( x) 0 lim [ f ( x) ⋅ g( x)]= lim has the indeterminate form . x→ x0 x→ x0 1 0 g( x) lim [ f ( x) ⋅ g( x)]= lim

x→ x0

70

x→ x0

g( x) ∞ 1 has the indeterminate form ∞ . f ( x) Algebra 10

EXAMPLE

30

Solution

1 Find lim( x ⋅ sin ). x →∞ x

The function has the indeterminate form ∞ ⋅ 0. We can transform it into the form

0 by writing 0

1 sin 1 x ). lim( x ⋅ sin ) = lim( x →∞ x →∞ 1 x x 1 1 = t. As x → ∞, → 0 so t → 0, x x 1 sin x ) = lim( sin t ) =1. and so lim( x →∞ t→0 1 t x

Let us take

EXAMPLE

31

Solution

Find lim

x →−∞

1 ⋅ (3x − 2). x +5

The indeterminate form is 0 ⋅ ∞. lim

x →−∞

1 3x − 2 ⋅ (3x − 2) = lim = 3. x→−∞ x +5 x +5

Check Yourself 12 Calculate the limits. x 9 1. lim( ⋅ sin ) x →∞ 3 x

2. lim( x ⋅ tan x →∞

1 2π x + 2) 2 ) 3. lim( x →∞ − 1) x ( x

4. lim x2 +5 x +1 ⋅ x →∞

1 x+ 4

Answers 1. 3

2. 2π

3. 0

4. 1

D. ∞ – ∞ AS A LIMIT Let f and g be two functions. Let x0 ∈ \, and let us assume that as x approaches x0, both f(x) and g(x) approach infinity, i.e. lim f ( x) = ∞ and lim g( x) = ∞.

x → x0

x → x0

Then lim [ f ( x) − g( x)] has the indeterminate form ∞ – ∞. We can find the limit of such x→ x0

functions by transforming the indeterminate form ∞ – ∞ into the indeterminate form

0 ∞ or , 0 ∞

as shown in the following examples. Limit of a Function

71

EXAMPLE

32

Solution

1 ⎞ ⎛ 2 Find lim ⎜ 2 − ⎟. x→1 ⎝ x − 1 x −1⎠

The indeterminate form is ∞ – ∞ as x approaches 1. We can rewrite the function to transform 0 the indeterminate form ∞ – ∞ into the indeterminate form : 0 ⎛ 1 ⎞ 2 1 ⎞ 2 – x–1 1– x ⎛ 2 lim ⎜ 2 – – = lim . ⎜ ⎟ = lim ⎟ = lim x →1 x – 1 x → x → x → 1 1 1 x – 1⎠ ( x – 1)( x+1) ( x – 1)( x+1) ⎝ ⎝ ( x – 1)( x +1) x – 1 ⎠ Now we have the indeterminate form lim x →1

EXAMPLE

33

Solution

0 and we can calculate the limit: 0

1– x –1 1 = lim =– . x → 1 ( x – 1)( x +1) ( x +1) 2

2 2 Find lim( 4 x – 6 x + 3 – 4 x + 3 x+1). x →∞

The indeterminate form is ∞ – ∞. Let us multiply and divide the expression by its conjugate. Then we get lim

( 4 x2 – 6 x + 3 − 4 x2 + 3 x +1)( 4 x2 – 6 x+ 3 + 4 x2 + 3 x+1)

x →∞

( 4 x2 – 6 x + 3 + 4 x2 + 3 x +1) = lim x →∞

4 x2 – 6 x + 3 – (4 x2 + 3 x +1) ⎛ 6 3 2 ⎛ + 2 ⎜⎜ 4x ⎜ 1 − x x 4 4 ⎝ ⎝

3 1 ⎞⎞ ⎞ 2 ⎛ ⎟ + 4 x ⎜1+ + 2 ⎟ ⎟⎟ 4 x 4x ⎠ ⎠ ⎠ ⎝

= lim x →∞

–9 x+ 2 2x + 2x

2 ⎞ ⎛ –9 x ⎜ 1 – ⎟ –9 x 9 9x ⎠ ⎝ = lim = lim =– . x →∞ x →∞ 4 x 2x+ 2x 4

EXAMPLE

34

Solution

2 Find lim( x – 2 x – 3 + x – 4). x→ – ∞

The indeterminate form is ∞ – ∞. Use the theorem: ⎛ –2 ⎞ lim( x2 – 2 x – 3 + x – 4) = lim ⎜ 1 ⋅ x+ + x – 4 ⎟= lim | x – 1|+ x – 4 x→ – ∞ 2 ⋅ 1 ⎝ ⎠ x→ – ∞

x→ – ∞

= lim(– x +1+ x – 4) = –3. x→ – ∞

72

Algebra 10

Check Yourself 13 Calculate the limits. 2 ⎞ ⎛ 1 + 1. lim ⎜ ⎟ x→1 x – 1 1 – x2 ⎠ ⎝

⎛ x2 x⎞ – ⎟ 2. lim ⎜ x→∞ 2 x – 5 2⎠ ⎝

3. lim ⎛⎜ 1 − 1 ⎞⎟ π cot x ⎠ x→ ⎝ cos x 2

4. lim( x2 + 3x +1 – x) 5. lim( x2 + 4 x +1 – x2 + 3 x + 3) x→∞

x →∞

6. lim( x + 2 – x – 2 ) 7. lim( x2 – 8 x +1+ x+ 4) x →∞

x →−∞

Answers 1 5 1. 2. 4 2

3. 0

4.

3 2

5.

1 2

6. 0

7. 8

∞

E. 1 AS A LIMIT Let f and g be two functions and let x0 ∈ \, and let us assume that as x approaches x0, f(x) approaches 0 and g(x) approaches infinity, i.e. lim f ( x) = 0 and lim g( x) = ∞.

x → x0

x → x0

Since lim(1+ f ( x)) approaches 1 but is not equal to 1, the limit lim(1+ f ( x)) g( x ) has the x → x0

x → x0

∞

∞

indeterminate form 1 . We can remove the indeterminate form 1 by using the following rule: Let lim f ( x) = 0, lim g( x) = ∞ and lim f ( x) ⋅ g( x) = k ( k ∈ \). Then x → x0

x → x0

x → x0

lim(1+ f ( x)) g( x ) = e k , where e ≅ 2.718.

x → x0

EXAMPLE

35

Solution

For f ( x) =

1 and g(x) = x, find lim(1+ f ( x)) g( x ). x →∞ x

We have lim f ( x) = lim x →∞

x→∞

1 = 0, x

lim g( x) = lim x = ∞ and x →∞

x →∞

⎛1 ⎞ lim f ( x) ⋅ g( x) = lim ⎜ ⋅ x ⎟ =1. x →∞ x→∞ x ⎝ ⎠

So k = 1 and by the given rule, x

1⎞ ⎛ lim(1+ f ( x)) g( x ) = lim ⎜1+ ⎟ = e1 = e. x→∞ x→∞ x⎠ ⎝ Limit of a Function

73

EXAMPLE

36

Solution

For f ( x) =

3 and g(x) = 5x, find lim(1+ f ( x)) g( x ). x →∞ x

We have lim f ( x) = lim x →∞

x→∞

3 = 0, x

For m, n ∈ \, lim(1+ x →∞

lim g( x) = lim 5 x = ∞ and x →∞

m nx ) = em⋅n. x

x →∞

⎛3 ⎞ lim f ( x) ⋅ g( x) = lim ⎜ ⋅ 5 x ⎟ =15. So k = 15 and by the given rule, x →∞ x →∞ x ⎝ ⎠ 5x

3⎞ ⎛ lim(1+ f ( x)) g( x ) = lim ⎜1+ ⎟ = e15. x→∞ x→∞ x⎠ ⎝

EXAMPLE

37

Solution

2 ⎞ ⎛ Find lim ⎜1+ ⎟ x →∞ 2 x +1 ⎠ ⎝

3 x+1

.

We can rewrite the limit in the form g(x) = 3x + 1. So f ( x) =

lim(1+ f ( x)) g( x ) using x →∞

f ( x) =

2 2 x +1

and

2 2 and lim f ( x) = lim = 0, x →∞ x→∞ 2 x +1 2 x +1

g( x) = 3x +1 and lim g( x) = lim 3 x+1= ∞ . x →∞

x →∞

This gives lim f ( x) ⋅ g( x) = lim x →∞

x→∞

2 6x+ 2 6 ⋅(3 x+1) = lim = = 3. So k = 3 and x →∞ 2x + 1 2 x +1 2

2 ⎞ ⎛ by the given rule, lim ⎜1+ ⎟ x →∞ 2 x +1 ⎠ ⎝

EXAMPLE

38

Solution

⎛ 3x +1 ⎞ Find lim ⎜ ⎟ x →∞ 3x – 1 ⎝ ⎠

We can write

3 x+1

= e3.

4x−2

.

3x +1 3x – 1+ 2 2 = =1+ . 3x – 1 3x – 1 3x – 1

⎛ 3x +1 ⎞ Then lim ⎜ ⎟ x →∞ 3x – 1 ⎝ ⎠

4 x− 2

4 x−2

2 ⎞ ⎛ = lim ⎜1+ ⎟ x→∞ x 3 –1⎠ ⎝

and we can take f ( x) =

2 and g(x) = 4x – 2. 3x – 1

8x – 4 8 8 ⎛ 2 ⎞ Since lim f ( x) ⋅ g( x) = lim ⎜ = , k= . ⎟ ⋅(4 x – 2) = lim x →∞ x →∞ 3x – 1 x →∞ 3 x – 1 3 3 ⎝ ⎠

⎛ 3x +1 ⎞ So lim ⎜ ⎟ x →∞ 3x – 1 ⎝ ⎠ 74

4 x− 2

4 x− 2

2 ⎞ ⎛ = lim ⎜1+ ⎟ x→∞ 3x – 1 ⎠ ⎝

8

= e 3. Algebra 10

Check Yourself 14 Calculate the limits. 5 1. lim ⎛⎜ 1+ ⎞⎟ x →∞ x⎠ ⎝

x

⎛ 3x – 2 ⎞ 4. lim ⎜ ⎟ x →∞ ⎝ 3x ⎠ ⎛ x+ 2 ⎞ 7. lim ⎜ ⎟ x →∞ x +5 ⎝ ⎠

x−2⎞ 2. lim ⎛⎜ ⎟ x →∞ ⎝ x ⎠ 4x

x

3 ⎞ ⎛ 5. lim ⎜ 1 – ⎟ x →∞ 2 x +7 ⎠ ⎝

⎛ 2 x +1 ⎞ 3. lim ⎜ ⎟ x →∞ ⎝ 2x ⎠ 2 x+5

x

2x – 5 ⎞ 6. lim ⎛⎜ ⎟ x →∞ 2 x + 3 ⎝ ⎠

x+ 4

3 x +1

Answers 1

1. e5

2. e–2

3. e 2

4. e

–

8 3

5. e–3

6. e–4

7. e–9

The figure shows an n-sided regular polygon inside a circle centered at O. 2π In the figure let |AB| = a, |OA| = r and m( ∠AOB) = so in the right triangle AOH, n a O a and a = 2rsin(∠AOH). sin( ∠AOH) = 2 = r 2r π r π Since ΔAOB is an isosceles triangle, m( ∠AOH ) = and a = 2 r sin . n n H π B A The perimeter of the polygon is therefore n ⋅ 2 r sin . a n As the number of sides of the regular polygon increases to infinity, the polygon gets closer and closer to a circle. π sin π n Therefore the circumference of the circle is lim n ⋅ 2 r sin = lim 2 r ⋅ π = 2 πr. n→∞ π n n→∞ n It can also be proved that the area of a circle is equal to πr2 by using a limit. (This is left as an exercise for you.)

Limit of a Function

75

EXERCISES A.

2 .2

0 as a Limit 0

| x |+2 x – 1 x → – ∞ 3x + 4| x |+5

g. lim

1. Calculate the limits. x–5 a. lim 2 x → 5 x – 25

x2 – x – 6 b. lim 2 x →–2 x + x – 2

x2 – 5 x – 6 c. lim x →–1 x +1

5 x3 – 5 d. lim x →1 x –1

x2 – 9 x – 5x +6

e. lim

2

x→ 3

2 – | x| x →–2 2+ x

g. lim

f. lim x→ 2

2 x – 3x + 4

x3 – m 3 x → m 2 x2 – mx – m 2

h. lim

sin5 x 7x

x→0

sin(3 x – 6) x → 2 tan( x – 2)

c. lim

sin x

e. lim

1 – cos 2 x

x→0

2 g. lim cos 2x – 1 x→0 x

i. lim x→0

sin2 x cos x – cos 2 x

b. lim

x–3 tan(2 x – 6)

d. lim

1 – cos 2 x sin 2 x

x→ 3

x→0

cos x +1 f. lim x →π ( π – x)2 π – 2x h. lim π sin 2 x x→ 2

2 3 j. lim (1 – x 2 )( x – 1) x →1 tan ( x – 1)

x2 +1 x→ – ∞ 5x – 4

d. lim

2005 x + 2010 x →∞ x2 +1

f. lim

c. lim

e. lim

76

x →∞

x2 + x +1 3 x3 + 4

x→ – ∞

x+ 2 4 – 2) x x

x4 + ax + 2 x →∞ ( x2 – 1)2 – a

x→0

b. lim( x→ – ∞

d. lim( x→1

2 x 3 – x2 – x) 1+ 2 x2

x 2 + ) x – 1 x2 – 4 x + 3

x →∞

b. lim( x2 + x +1 – x2 + 2 x – 3) x →∞

c. lim( x2 + 2 x + 2 – x+1) x →∞

d. lim( x x2 +1 – x2 ) x →∞

E. 1∞ as a Limit 7. Calculate the limits.

x+ 3 x ) x

3 b. lim(1 – )3 x x →∞ x

d. lim(

6 x +1 2 x ) 6x

x →∞

e. lim(1 –

2 2 x+1 ) 3x +1

f. lim(

5x +1 3 x+1 ) 5x – 1

g. lim(1+

7 )2 x 5x + 3

h. lim(

x + 2 9 x –1 ) x+ 4

x →∞

x2 + x + 2 3+ 5 x5 +1

d. lim(sin 2 x ⋅ cot 5 x)

a. lim( x2 +5 x +1 − x)

x →∞

b. lim

5 x+ 2

6. Calculate the limits.

c. lim(

3x3 + 2 x2 +1 x →∞ x3 – 1

x →∞

5. Calculate the limits.

2 a. lim(1+ )x x →∞ x

3. Calculate the limits.

b. lim x2 – x + 3 ⋅

D. ∞ – ∞ as a Limit

x→0

B. ∞ as a Limit ∞ a. lim

c. lim( 3x ⋅ tan 3 ) x →∞ x

c. lim(

2. Calculate the limits. a. lim

6 ) 7x

2 x a. lim( x − ) x→∞ 3x + 4 3

2 2 x+4 – 16 i. lim x+1 x→ 0 2 –2

x5 – 1

4. Calculate the limits. x →∞

2

5

x→ – ∞

C. 0 ⋅ ∞ as a Limit

a. lim( x ⋅ sin

x–2

x2 + x – 2 + 3 x

h. lim

x→∞

i. lim( x →∞

3x +1 3 x+1 ) 3x + 2

x →∞

x →∞

j. lim( x →∞

2 x +7 x –1 ) 2x – 1 Algebra 10

A. CONTINUITY AT A POINT In this section we will look at the concept of continuity of a function. During our study of the limit of f(x) as x approaches x0, we have continuously emphasized that the limit is not necessarily equal to f(x0). Indeed, what actually happens at the point x0 is not important for the limit of the function at that point. However, the nature of f(x0) becomes important when we are considering the continuity of a function.

Definition

continuity at a point, discontinuity at a point Let A be a subset of \, let f: A → \ be a function and let x0 ∈ A. If lim f ( x) = f ( xo )

x → xo

then we say f is continuous at x0. Otherwise, f is said to be discontinuous at x0. It is important to note that for a function f to be continuous at x0, three things are necessary: 1. The limit of f(x) as x → x0 must exist. 2. The function f must be defined at the number x0, i.e. f(x0) must exist. And also, by the definition above, 3. lim f ( x) = f ( xo ). x→ xo

If even one of these three conditions is not satisfied, the function f is said to be discontinuous at point x0. Let us begin by looking at the continuity of some functions at given points.

y y = f( x )

The figure shows the graph y = f(x). At x = a, lim f ( x) does not exist.

f(a)

x→ a

So f is discontinuous at x = a. O

Limit of a Function

a

x

77

y

At x = b, lim f ( x) = L but f(b) does not exist. x→ b

This means that f is not defined at x = b. So f is discontinuous at point b.

y = f(x) L O

x

b

y

At x = c, lim f ( x) = L and f(c) exists, but x→ c

lim f ( x) ≠ f ( c). x→ c

So f is discontinuous at point c.

f(c) y = f(x) L x

c

O

y

At x = d, lim f ( x) = f ( d) .

y = f(x)

x→ d

So f is continuous at point d. f(d) = L d

O

x

Note The graph of a function which is continuous at a point has no gaps or breaks in its line at that point. In other words, we can draw the graph of the function without lifting our pen from the paper when we pass through the point. EXAMPLE

39

Examine the continuity of the function f(x) in the figure at the points x = –3, x = –1, x = 2 and x = 4.

y 7 6 5 4 3 2 1 –4 –3 –2 –1 O

78

y = f(x)

1 2 3 4

x

Algebra 10

Solution

a. The graph shows f(–3) = 2 and lim f ( x) = 2, so x →−3

lim f ( x) = f ( −3). So f is continuous at x = –3.

x→−3

b. f(–1) is not defined. So f is discontinuous at x = –1. c. lim f ( x) does not exist. So f is discontinuous at x = 2. x→2

d. lim f ( x) = 5 but f(4) = 6. x→4

Since lim f ( x) ≠ f (4), f is discontinuous at x = 4. x→4

EXAMPLE

40

Solution

Examine the continuity of f: \ → \, ⎧ x2 +1 if x < 1 ⎪⎪ f ( x) = ⎨ 2x if x = 1 at the point x =1. ⎪ ⎩⎪ − x + 3 if x > 1

Since f(1)= 2 ⋅ 1 =2, f(x) is defined at x = 1. Let us find the limit of the function at x = 1:

3

lim− f ( x) = lim( x2 +1) =12 +1= 2 and −

x →1

y

y = x2 + 1

2

x →1

lim f ( x) = lim( −x + 3) = −1+ 3 = 2, so +

x →1+

x →1

1

lim− f ( x) = lim+ f ( x) = 2 and so lim f( x) = 2.

x →1

x →1

x →1

Since lim f ( x) = f (1), f is continuous at x = 1.

–1

O

x→1

EXAMPLE

41

Solution

1

2

3

x y = –x + 3

Examine the continuity of f: \ – {–2} → \, ⎧ x3 − 8 if x ≠ 2 and x ≠ –2 ⎪ 2 f ( x) = ⎨ x − 4 at the points x = 2 and x = −2. ⎪ if x = 2 ⎩ 3

a. At x = 2, f(2) = 3. lim f ( x) = lim x→ 2

x→ 2

= lim x→ 2

x3 − 8 ( x − 2)( x2 + 2 x + 4) = lim x2 − 4 x → 2 ( x − 2)( x + 2) x2 + 2 x + 4 12 = =3 x+ 2 4

Since lim f ( x) = f (2), f is continuous at x = 2. x→ 2

b. At x = –2, f(–2) is not defined, so f is discontinuous at x = –2. Limit of a Function

79

Check Yourself 15

y 4 3 2

1. Examine the continuity of the function f shown in the figure at the points x = –3, x = –2, x = –1, x = 0, x = 2 and x = 3.

y = f(x)

1 –4

–2 –1 O

1 2 3 4

x

2. Examine the continuity of f: \ → \, ⎧ ⎪ ⎪ f(x) = ⎨ ⎪ ⎪ ⎩

8x − 3 if x < −3 3 3x x−6

if x = −3 at the point x = – 3. if x > −3

⎧⎪ x2 + 5x if x < −1 3. Examine the continuity of f: \ → \, f ( x) = ⎨ at x = −1. 3 ⎩⎪ x − 4x if x ≥ −1

⎧ ⎪ ⎪⎪ 4. Examine the continuity of f: \ → \, f(x) = ⎨ ⎪ ⎪ ⎪⎩ 2 ⎧ mx + nx if ⎪ 8 if ⎪⎪ 5. f: \ → \, f(x) = ⎨ ⎪ mx − n if ⎪ 5+ x2 ⎪⎩

x−9 x−3 3

if x < 0 if x = 0 at x = 0.

3 1+ x2

if x > 0

x < −2 x = −2

is a function which is continuous at x = –2.

x > −2

Find the values of m and n. Answers 2. continuous

80

3. discontinuous

4. continuous

5. m = –5, n = –14

Algebra 10

EXERCISES

2 .3

A. Continuity at a Point 1. The figure shows the graph

5. Examine the continuity of f: \ → \, y 5

of a function f. Examine the continuity of f at the integer points in the domain.

4 y = f(x)

3 2

⎧ 2 x2 +1 if x < 1 ⎪ f ( x) = ⎨ x + 2 if x =1 at the point x= 1. ⎪ ⎩ − x + 4 if x > 1

1 – 7 –3 –2 –1 2

O

1

2

37 2

x

⎧ x2 + 2 x + 2 if x < −1 ⎪ m if x = −1 6. The function f ( x) = ⎨ ⎪ ⎩ mx + n if x > −1

2. Examine the continuity of f: \ → \,

is continuous at x = –1. Find m + n.

⎧⎪ x + 4 if x = 4 at the point x = 4. f ( x) = ⎨ ⎪⎩ 2 x if x ≠ 4

3. Examine the continuity of f: \ → \, ⎧⎪ x +6 if x > 3 f ( x) = ⎨ at the point x = 3. 2 ⎪⎩ x if x < 3

4. Find the points at which the function ⎧ ⎪ f ( x) = ⎨ ⎪ ⎩ Limit of a Function

⎧ ⎪ 7. The function f( x) = ⎨ ⎪ ⎩

mx + 3 if x > 2 x+1

if x= 2

n2 − 1

if x < 2

is continuous at x = 2. Find the possible values of m and n.

2 2 ⎪⎧ ax − a x + 3 if x < −1

8. f ( x) = ⎨

2 if 1 ≤ x ≤ 2 is discontinuous.

is given. 3 2 ⎪⎩ a x + ax +5 if x ≥ −1 Find the possible values of a if f is continuous at

1 if x > 2

the point x = –1.

1 if x < 1

81

B. Continuity on an Interval 9. Find the largest interval on which each function is continuous. a. f(x) = x3 – x2 + x – 1 c. f ( x) =

1 x −1

e. f ( x) = x3 + g.

1 x

b. f ( x) =

x x +2

d. f ( x) =

x+ 2 x − 5x − 6

f.

2

2

f ( x) = 3 2 x − 5

f ( x) = 3 3 x + 4 + x2 − 4

h. f ( x) =

x2 − 2 x +1 x2 − 3 x − 4

10. Find the largest interval on which each function is continuous. 1 a. f ( x) = cos x 2

b. f ( x) =

c. f ( x) = sin x

d. f ( x) = sin(

e. f(x) = arccos(x – 4)

f. f ( x) = sin(log

g.

f ( x) = arcsin(

sin x 1 − cos x x+ 2 ) x−2 x+2 ) x−2

3x + 4 ) x −1

⎧ x3 − 27 if x > 3 ⎪ 3−x ⎪ if x = 3 a 11. f ( x) = ⎨⎪ ⎪ x2 − bx + 27 if x < 3 ⎩

is a continuous function in the set of real numbers. Find a + b. ⎧ ⎪

12. f ( x) = ⎨

ax x2 +1

if x > 1

⎪ ax − 2 if x ≤ 1 ⎩ is continuous in \. Find a.

82

Algebra 10

CHAPTER REVIEW TEST 1. The figure shows the graph of f(x). At which integer point in the interval (1, 7) does a finite limit of f(x) exist?

2

y

5. Find xlim →0 −

y = f(x) 3 2 1 O

A) 0 1 2 3 4 5

6

7

3x +| x| . 3x + 2| x|

B) 1

C) 2

D) 3

E) 4

C) e5

D) e–15

E) e15

C) 6

D) 8

E) 16

x

–3

A) 2

B) 3

C) 4

D) 5

E) 6

5 x

6. Find lim(1+ ) 3 x. x →∞

A) e–3

B) e3

2. Find lim( x7 + x6 − x5 ). x →−∞

B) ∞

A) –∞

C) 0

D) 2

E) –1

7. Find lim( x→0

1 x

. 3. Find xlim →0 –

A) 0

B) 1

A) 2 C) –1

D) ∞

B) –∞

Chapter Review Test 2

B) 4

E) –∞

x2 – 3 x+ 4 – x). 8. Find lim( x→∞

2 x 3 – x2 – 1 4. Find lim 2 . x →−∞ x + 2 x +1

A) ∞

sin 4 x x+2 ) . x

C) 2

D) –2

E) 0

A) 1 2

B) – 3 2

C) 3 2

D) – 1 2

E) 0 83

9. Find lim( x→ a A)

3a 2

a 3 − x3 ). a 2 − x2

B) a 2

13. Find lim x →∞ C) 3 2

D) 3a

A) –1

B) 1

C) 2

1 2a

A)

6 5

B) 2 5

4 3 2 1 O –1 –2

y = f(x)

11. Find lim x→9 A) 0

x–3

B) 3

x →∞

1 2

3 4

D) 3

5

x

D) 9

12 x3 + 2 x2 − 1 . x→∞ x3 +1

84

B) –6

C) 12

E)

3 5

x

A) e5

B) e10

15. Find lim

sin3 2 x . x3

A) 0

B) 1

C) e2

D) e

E) 1

C) 2

D) 4

E) 8

E) ∞

3 16. Find lim( x −4 8 x +8 ).

12. Find lim A) 6

D) – 1 5

E) 4

.

C) 6

1 4

14. Find lim( x + 5 )2 x.

x→0

x–9

C) –

y

10. The graph of a function f(x) is shown in the figure. Find the sum of the left-hand limits of the function at the integer values in the interval (0, 5].

E)

( x + 2)(3x − 1) . 5 x2 − 4

x→ 2

D) –12

E) 14

A) 0

x − 4x

B) 2

C) 4

D) 8

E) 16 Algebra 10

A. ANGLES AND THE UNIT CIRCLE In your previous studies you have already learned the basic principles of trigonometry on the unit circle and in right triangles. Before we begin our study of trigonometry, it will be helpful to review these basic concepts and definitions.

Definition

angle, directed angle An angle is the union of two rays which have the same initial point. If one of the rays of an angle is called the initial side of the angle and other ray is called the terminal side, then the angle is called a directed angle.

Definition

negative angle, positive angle If a directed angle is measured in a clockwise direction from its initial side then the angle is a negative angle. If the angle is measured in a counterclockwise direction then it is a positive angle. In trigonometry we use both positive and negative angles.

l itia

in ter

e

na mi

ter

a

min

de

l si

sid

a init ial s

al s

ide

ide

negative angle: a = – 60°

positive angle: a = 60°

We can measure angles using different units of measurement. The most common units are degrees and radians. We write ° to show a degree measurement: one full circle measures 360°. We write R to to show a radian measurement: one full circle measures 2πR. We can also 3π omit the radian notation if it is clear that an angle is in radians: α = means that the angle 2 3π α measures radians. 2 We can use a simple formula to convert between degree (D) and radian (R) measures: D R = 180 ° π π π , 45° = , ... . 2 4 In trigonometry we often work with angles drawn in the coordinate plane.

For example, 360° =2πR, 90° =

86

Algebra 10

unit circle, quadrant

Definition

The circle whose center lies at the origin of the coordinate plane and whose radius is 1 unit is called the unit circle. The coordinate axes divide the unit circle into four parts, called quadrants. The quadrants are numbered in a counterclockwise direction. in degrees:

180°

in radians:

90°

2nd quadrant

1st quadrant

3rd quadrant

4th quadrant

0º p

360°

p 2

2nd quadrant

1st quadrant

3rd quadrant

4th quadrant

0 2p

3p 2

270°

quadrantal angles

Definition

The intersection points of the unit circle and the coordinate axes correspond to angles measured on the circle. These angles are called quadrantal angles. In other words, 0°, 90°, π 3π 180°, 270°, 360°, ... and 0, , π, , 2π, ... are quadrantal angles. 2 2 π⎞ ⎛π ⎞ ⎛ In the unit circle, if α ∈ ⎜ 0, ⎟ then α is in the first quadrant, if α ∈ ⎜ , π ⎟ then it is in the 2 ⎝2 ⎠ ⎝ ⎠ 3π ⎞ ⎛ ⎛ 3π ⎞ second quadrant, if α ∈ ⎜ π, ⎟ then it is in the third quadrant and if α ∈ ⎜ , 2π ⎟ then it 2 ⎠ ⎝ ⎝ 2 ⎠ is in the fourth quadrant. The same applies to the equivalent intervals in degrees. EXAMPLE

1

Solution

In which quadrant does each angle lie? a. 75°

b. 228°

c. 305°

d. 740°

e. –442°

f.

7π 3

g –

17 π 5

a. 75° < 90°, so it is in the first quadrant. b. 228 ° ∈ (180 °, 270 °), so it is in the third quadrant.

We write ≡ to show that two angles are coterminal: α ≡ β means that α and β are coterminal. Be careful! α ≡ β does not mean α = β. For example, 740° ≡ 20° but 740° ≠ 20°.

c. 305 ° ∈ (270 °, 360 °), so it is in the fourth quadrant. d. 740° = 20° + (2 ⋅ 360°) ≡ 20° and 20° ∈ (0 °, 90 °). So 740° is in the first quadrant. e. –442° = 278° – (2 ⋅ 360°) ≡ 278° and 278 ° ∈ (270 °, 360 °). So – 442° is in the fourth quadrant. f.

7π π π π ⎛ π⎞ 7π is in the first quadrant. = + 2 π ≡ and ∈ ⎜ 0, ⎟, so 3 3 3 3 ⎝ 2⎠ 3

g. –

3π ⎛ π ⎞ 17 π 17 π 3π 3π and is in the second quadrant. ∈ ⎜ , π ⎟ , so – = − (2 ⋅ 2 π) ≡ 5 5 5 5 ⎝2 ⎠ 5

Trigonometry: Functions, Equations and Inequalities

87

Remember that there is a correspondence between the points on the unit circle and the real number line. Imagine the real number line placed vertically next to the unit circle, so that O on the real number line coincides with A(1, 0) on the unit circle. Then we can ‘wrap’ the real number line around the top and bottom halves of the circle to see the correspondences: in the figure, we can see that point D corresponds to π ≅ 1.571 on the number line, 2 point F corresponds to − π , point B(a, b) 2 corresponds to x1, and so on. The intercept of one radian and the unit circle corresponds to 1 on the real number line.

EXAMPLE

2

Solution

E

D

p 2 x2 x1

C

b

B(a, b)

A(1, 0) a

F

G x3 –

p 2

Find the real number which corresponds to the central angle 60° on the unit circle, using π ≅ 3.14. From the formula R≅

60° D R R 60° ⋅ π π we have = = , so R = = . Using π ≅ 3.14 gives us 180° 3 180 ° π 180 ° π

3.14 ≅ 1.047. This is the required real number. 3

Check Yourself 1 1. Convert the measures to radians. a. 30°

b. 135°

c. 210°

2. Convert the measures to degrees. 5π π a. b. c. 5π 6 3 4 Answers π 3π 7π 1. a. b. c. 6 4 6 2. a. 60° b. 150° c. 225° 88

d. 900° d. 10π

d. 5π d. 1800° Algebra 10

B. PRIMARY DIRECTED ANGLES 1. Coterminal Angles Definition

standard position of an angle An angle in the coordinate plane whose vertex is at the origin and whose initial side lies along the positive x-axis is said to be in standard position.

Definition

coterminal angles Two or more angles whose terminal sides coincide with each other when they are in standard position are called coterminal angles. Let us look at an example of coterminal angles. The figure shows a unit circle. The positive angle ∠AOP corresponds to the arc AùEP and the negative angle ∠AOP corresponds to the arc AùFP. These angles are coterminal. The measure of the positive angle ∠AOP is m(∠AOP) = α ° and the measure of the negative angle ∠AOP is m(∠AOP) = –(360 – α)°.

y P F E q

a O

We can also express the measure of each angle in radians. Since this is a unit circle, if the length of the arc AùEP is θ then the measure of the positive angle ∠AOP is m(∠AOP) = θ and the measure of the negative angle ∠AOP is m(∠AOP) = –(2π – θ ).

x

A

Now assume that point P in the figure is moving around the circumference of the unit circle from point A in the counterclockwise direction. Study the following table.

Position of point P (moving counterclockwise)

Measure of the central angle for AïP Degrees

Radians

P lies on the positive x-axis

0°

0

P lies on the positive y-axis

90°

π 0

P lies on the negative x-axis

180°

π

P lies on the negative y-axis

270°

3π 2

P lies on the positive x-axis after one complete revolution

360°

2π

P lies on the ray [OP after one complete revolution

360° + α

2π + θ

P lies on the ray [OP after a second revolution

720° + α

4π + θ

k ⋅ 360° + α

2kπ + θ

P lies on the ray [OP after its kth revolution

Trigonometry: Functions, Equations and Inequalities

89

EXAMPLE

3

Solution

For each angle, write the set of coterminal angles with the same unit of measurement. 5π a. 175° b. 4 Coterminal angles differ by an integral multiple of complete angles. a. {175° + k ⋅ 360°, k ∈ ]} = {...,–545°, –185°, 175°, 535°, 895°,...} b. {

5π 11π 3 π 5 π 13 π 21 π + k ⋅ 2π, k ∈ ]} = {..., − , − , , , ,...} 4 4 4 4 4 4

Note

a.

The angles in part a and part b are coterminal. Therefore, if we graph them in standard position, these angles will have the same terminal side.

EXAMPLE

4

Solution

5

Solution

5p ––– 4

175° –185°

3p – ––– 4

Find the arc length which corresponds to the central angle 40° on the unit circle (π ≅ 3). 40° R 2π D R we have = = , so R = . We know that on the unit circle, the radian 180 π 180 π 9 measure of a directed angle is equal to the length of the directed arc corresponding to the

Since

angle. So the arc length is

EXAMPLE

b.

2π 2π 2 ⋅ 3 2 , and using π ≅ 3 gives us ≅ = ≅ 0.6. 9 9 9 3

π Find the coordinates of the terminal point of the arc with length which is in standard 2 position on the unit circle.

The circumference of a unit circle measures 2π.

y

π So represents a quarter of the circle. 2 π Therefore the arc length corresponds to the 2

point B(0, 1) on the unit circle.

B(0, 1)

A¢(–1, 0)

A(1, 0)

x

Furthermore, the arc length π corresponds to the point A′(–1, 0) on the unit circle and corresponds to B′(0, –1). 90

3π 2

B¢(0, –1)

Algebra 10

2. Primary Directed Angles and Arcs primary directed angle

Definition

Let β be an angle which is greater than 360°. Then the positive angle α ∈ [0, 360°) which is coterminal with β is called the primary directed angle of β.

In order to find a primary directed angle α we must divide the initial angle by 360. We must not simplify before the division, because 360 represents a complete rotation. For example, the remainder in the operation 5000÷360 gives us the required primary directed angle whereas the simplified version 500÷36 does not.

In other words, the primary directed angle of β is the smallest positive angle that is coterminal with β. If we divide β by 360°, the remainder will be the primary directed angle. m(β ) = k ⋅ (360°) + m(α), k ∈ ]. For example, 30° is the primary directed angle of 390° because 390° = 1 ⋅ 360° + 30°.

We know that the radian measure of any angle is equal to the length of the arc which corresponds to its central angle in the unit circle. The circumference of a unit circle is 2π. Therefore any two real numbers that differ by integral multiples of 2π will coincide at the same point on the circle. primary directed arc

Definition

The positive real number t ∈ [0, 2π) which differs from a real number by integral multiples of 2π is called a primary directed arc. Since t is the smallest positive real number that is coterminal with a given angle θ, we can find t by subtracting integral multiples of complete rotations from θ, or alternatively by dividing θ by 2π and considering the remainder:

θ = k ⋅ (2π) + t, k ∈ ].

EXAMPLE

6

Solution

Find the primary directed angle of each angle, using the same unit. 75π 75π a. 7320° b. –7320° c. d. − 8 8 a.

7320 – 7200 120

360 20 number of rotations

Trigonometry: Functions, Equations and Inequalities

91

y

y

x 120° x

21st rotation 20 complete rotations in the positive direction

7320° = (20 ⋅ 360°) + 120°, so 120° is coterminal with 7320°. So the primary directed angle of 7320° is 120°. b. Solution 1 7320° = (20 ⋅ 360°) + 120° –7320° = –(20 ⋅ 360° + 120°) = (–20 ⋅ 360°) – 120° = –120° – 120° ≡ 240° (coterminal angles) – 7320° ≡ 240° (coterminal angles) Solution 2 –7320° = (–21) ⋅ 360° + 240° Therefore the primary directed angle of –7320° is 240°. y

y

+240° x

x –120°

20 complete rotations in the negative direction

92

21st rotation

Algebra 10

c. Solution 1

y

75π 64 π 11π = + 8 8 8 75π 11π = 4 ⋅ (2 π)+ 8 8

number of rotations

α=

x

11π 8

Solution 2 1. Divide the numerator by twice the denominator: 2 ⋅ 8 = 16 and 75 ÷ 16 = (4 ⋅ 16) + 11.

75p 8

2. Multiply the remainder by π π 11π : 11 ⋅ = . denominator 8 8 So the primary directed angle of

5th rotation

75 16

11π 75π is . 8 8

– 64 4 11

number of rotations

remainder d. If the angle was positive, the remainder 11π would be as we found in part c. But the 8 11π angle is negative, so the remainder is – . 8

– 75p 8

y

Because a coterminal angle must be positive,

x

11π 11π 5 π = . ≡ 2π – 8 8 8 5π So the primary directed angle is . 8

we calculate −

EXAMPLE

7

Solution

Find the primary directed angle of θ = −30° 42′ 15′′. The primary directed angle must be positive, so we need to find the positive difference from 360°. Let the primary directed angle be θ ′. To make the calculation easier we can write 360° as 359° 59′ 60′′. Then

1° = 60′ and 1′ = 60′′ so 360° = 359° 59′ 60′′.

359° 59′ 60′′ – 30° 42′ 15′′ ––––––––––––––––––––– 329° 17′ 45′′. So θ′ = 329° 17′ 45′′.

Trigonometry: Functions, Equations and Inequalities

93

Check Yourself 2 1. Find the primary directed angle of each angle, using the same unit of measurement. 3π 33π a. 100° b. 7200° c. d. 2 5 11π 5π e. –400° f. –50° g. − h. − 3 4 2. For each angle, write the set of coterminal angles with the same unit of measurement. π 3π a. 30° b. 120° c. d. 3 2 Answers 3π 3π π 3π 1. a. 100° b. 0° c. d. e. 320° f. 310° g. h. 2 3 5 4 2. a. {..., –690°, –330°, 30°, 390°, ...} b. {..., –600°, –240°, 120°, 480°, 840, ...} 11π 5 π π 7 π 13 π 5π π 3π 7 π 11π , – , , , , ...} , , , ...} c. {..., – d. {..., – , – , 3 3 3 3 3 2 2 2 2 2

C. TRIGONOMETRIC RATIOS 1. Definition C hypotenuse (hyp)

opposite side (opp)

q adjacent side (adj)

A

B

Consider the right triangle in the figure. The table shows the trigonometric ratios for the acute angle θ.

94

Ratio name

Ratio abbreviation

Ratio definition

Abbreviated definition

sine

sin θ

length of side opposite θ length of hypotenuse

opp hyp

cosine

cos θ

length of side adjacent to θ length of hypotenuse

adj hyp

tangent

tan θ

length of side opposite θ length of side adjacent to θ

opp adj

cotangent

cot θ

length of side adjacent to θ length of side opposite θ

adj opp

secant

sec θ

length of hypotenuse length of side adjacent to θ

hyp adj

cosecant

csc θ

length of hypotenuse length of side opposite θ

hyp opp

Algebra 10

EXAMPLE

8

C

The figure shows a right triangle. Write the six trigonometric ratios for the angle θ.

8

10

q A

Solution

6

sin θ =

opp 4 = hyp 5

cos θ =

adj 3 = hyp 5

tan θ =

opp 4 = adj 3

cot θ =

adj 3 = opp 4

sec θ =

hyp 5 = adj 3

csc θ =

hyp 5 = opp 4

B

We know that any ratio can be expanded or simplified by multiplying its numerator and denominator by the same non-zero number. For example: 2 2k 4 40 200 = = = = =... etc. where k is any non-zero number. 3 3k 6 60 300

This property is also used in trigonometry. Look at the two right triangles below. C¢

C

b¢

b

a

a¢

q B

q

A

c

B¢

sin θ =

a b

c¢

sin θ =

A¢

a′ b′

Although the lengths of the sides of the triangles are different, the two trigonometric ratios a a′ for the common angle θ are the same: = . In other words, the sides are in proportion. We b b′ say that these triangles are similar.

Trigonometry: Functions, Equations and Inequalities

95

EXAMPLE

9

Two right triangles are shown below. Find the trigonometric ratios for the angle α in each triangle and show that they are equal. C

C 13

5

a 12

A

B

26 10

a 24

A

opp 5 = hyp 13 adj 12 cos α = = hyp 13 opp 5 tan α = = adj 12 adj 12 cot α = = opp 5

Solution sin α =

B

opp 10 5 = = hyp 26 13 adj 24 12 cos α = = = hyp 26 13 opp 10 5 tan α = = = adj 24 12 adj 24 12 cot α = = = opp 10 5 sin α =

The ratios are the same because the sides are in proportion: these are similar triangles. EXAMPLE

10

2 In a right triangle, θ is an acute angle such that cos θ = . Find the sine, tangent and 3 cotangent ratios of the same angle.

Solution We do not know the lengths of the sides of the triangle. However, we know that any right triangle with angle θ will be similar to this triangle. So we can use the numer2 ator and denominator of the given ratio ( ) as two sides 3 of the triangle, as shown in the figure. Now we can use the Pythagorean Theorem to find the length of the opposite side: opp2 + 22 = 32 2

opp + 4 = 9

hyp = 3 opp

q 2 cos q = 3

adj = 2

opp2 = 5 so opp = 5.

The resulting right triangle gives us the following results: sin θ = 96

opp 5 opp 5 adj 2 = ; tan θ = = ; cot θ = = . hyp 3 adj 2 opp 5 Algebra 10

Check Yourself 3 1. In a right triangle, θ is an acute angle such that tan θ = 4. Find the sine, cosine and cotangent ratios of the same angle. 2. One leg of an isosceles right triangle is 1 unit long. Find all the trigonometric ratios of one of the two equal acute angles in the triangle. Answers 4 1 1 1. sin θ = , cos θ = , cot θ = 4 17 17 2. sin 45° = cos 45° =

1 2

, tan 45° = cot 45° =1, csc 45° = sec 45° = 2

2. Special Triangles and Ratios Certain right triangles have ratios which we can calculate easily using the Pythagorean Theorem. One example is the isosceles right triangle which we obtain when we bisect a square diagonally. If the square has side length 1 unit we obtain the isosceles right triangle shown in the figure.

1 45° 1

®

1

ñ2

1

45° 1

1

A

Another example is the right triangle which we obtain when we bisect of an equilateral triangle from an altitude to a base. If the triangle has side length 2 units we obtain the right triangle shown in the figure.

A

A

60°

2

2

30° 30°

2

30°

2

2

ñ3 60°

B

60°

60°

C

2

B

1

60°

60°

H

C

1

ñ3

B

1

H

In each example we can find the unknown length using the Pythagorean Theorem: 12 + 12 = 2 and

2 2 – 12 = 3. We can use these two special right triangles to make a table of trigonometric ratios for some common angles. θ in degrees

θ in radians

sin θ

cos θ

tan θ

cot θ

30°

π 6

1 2

3 2

1 3

ñ3

45°

π 4

1 2

1 2

1

1

60°

π 3

3 2

1 2

ñ3

1 3

Trigonometry: Functions, Equations and Inequalities

97

EXAMPLE

11 B

A surveyor located on level ground at a point A is standing 36 m from the base B of a flagpole. The angle of elevation between the ground and the top of the pole is 30°. Find the approximate height h of the flagpole.

A

angle of elevation A

h 30°

B

angle of depression

Solution

36 m

We know that tan 30 ° =

1 3

from the trigonometric table we have just seen.

Looking at the figure, we also know that tan 30 ° = Therefore

EXAMPLE

12

opp h = . adj 36

1 36 h ≅ 20.78 m. = , and so h = 36 3 3

In the figure, m(∠BAD) = m(∠DAC) = 30° and BD = 12 units. Find the value of x.

A

30°

30°

B

12

D

C x

Solution

In the right triangle ABD, tan 30 ° =

1 3

=

In the right triangle ABC, tan 60 ° = 3 =

12 so AB = 12ñ3. AB 12+ x 12 3

so 12 + x = 12ñ3 ⋅ ñ3 = 36.

So x = 36 – 12 = 24 units. 98

Algebra 10

EXAMPLE

13

In the figure, m(∠ABC) = 30°, m(∠ACB) = 45° and AB = 6 units. Find the value of x.

A 6 45°

30° B

H

C

x

Solution

In the right triangle ABH, sin 30 ° =

1 AH = so AH = 3. 2 6

3 BH = so BH = 3ñ3. Since angle H is a right angle 2 6 and angle C measures 45° then in the triangle AHC, m(∠A) is also 45°. Therefore AHC is an

Also, in the same triangle cos 30 ° =

isosceles right triangle. So AH = HC = 3. Since BC = BH + HC, we have BC = x = 3ñ3 + 3.

Check Yourself 4 1. Find the length x in each triangle. a.

b.

A

A

c.

A

15°

8

30°

30°

2ñ2 60°

45°

B H C 1444442444443 x

2. Solve for x:

B

30° 14243 D x

C

B

8

D

C

x

tan 30° ⋅ csc60° = x ⋅ cot 30° sin 45°. cos 45° ⋅ sin60°

Answers 1. a. 4( 2 + 6 ) 3

Trigonometry: Functions, Equations and Inequalities

b. 2ñ3 – 2

c. 16

2.

8 9

99

D. TRIGONOMETRIC IDENTITIES The trigonometric ratios are related to each other by equations called trigonometric identities.

1. Basic Identities a. Pythagorean identities Property

Pythagorean identities For all θ ∈ \, 1. sin2θ + cos2θ = 1 2. tan2θ + 1 = sec2θ 3. cot2θ + 1 = csc2θ .

C hypotenuse (hyp)

q A

Proof

1. By the Pythagorean Theorem we have opp2 + adj2 = hyp2. 2

Therefore, Be careful! sin2 θ = sin θ ⋅ sin θ and cos2 θ = cos θ ⋅ cos θ, etc. We do not write sinθ 2 because it is not clear what we mean: sin (θ 2) or (sin θ )2?

2

2

opp + adj hyp = 2 hyp hyp 2

adjacent side (adj)

B

(1)

(divide both sides by hyp 2)

hyp 2 opp2 adj 2 = + hyp2 hyp 2 hyp 2 opp 2 adj 2 ) +( ) =1. hyp hyp adj opp , we have sin2θ + cos2θ = 1. Since sin θ = and cos θ = hyp hyp opp2 + adj 2 hyp 2 2. Dividing both sides of (1) by adj2 gives = , i.e. adj2 adj 2 2 2 2 opp adj hyp + = , adj2 adj 2 adj 2 (

(

2 hyp 2 opp 2 adj =( ). ) + 2 adj adj adj

Since tan θ =

opp hyp and sec θ = we have tan2θ + 1 = sec2θ. adj adj

3. Dividing both sides of (1) by opp2 gives So

opp2 adj 2 hyp 2 + = , opp2 opp 2 opp 2

opp2 opp

2

+(

opp2 + adj 2 hyp 2 = . opp2 opp 2

adj 2 hyp 2 ) =( ). opp opp

Since cot θ = 100

opposite side (opp)

hyp adj and csc θ = we have 1 + cot2θ = csc2θ , i.e. cot2θ + 1 = csc2θ. opp opp Algebra 10

b. Tangent and cotangent identities Property

tangent and cotangent identities sin θ cos θ cos θ 2. cot θ = sin θ 3. tan θ ⋅ cot θ = 1

1. tan θ =

Proof

opp hyp opp opp sin θ adj 1. We know that sin θ = and cos θ = , so = . = adj hyp hyp adj cos θ hyp sin θ = tan θ . So cos θ adj hyp cos θ adj cos θ = = , i.e. 2. Similarly, = cot θ . opp sin θ opp sin θ hyp

3. Consequently,

sin θ cos θ ⋅ = 1. So tan θ ⋅ cot θ = 1. cos θ sin θ

c. Reciprocal identities Property

Proof

Remember! csc θ =

1 sin θ

1 cos θ (the first letters of reciprocal ratios are opposite: 1 1 c= ,s= ) c s sec θ =

reciprocal identities 1. csc θ =

1 sin θ

2. sec θ =

1 cos θ

1. We know that sin θ =

Since csc θ =

1 hyp , csc θ = . opp sin θ

2. Similarly, cos θ =

Since sec θ =

opp 1 1 hyp , so = = . opp opp hyp sin θ hyp

adj 1 1 hyp so = = . adj hyp cos θ adj hyp

hyp 1 , sec θ = . adj cos θ

Trigonometry: Functions, Equations and Inequalities

101

EXAMPLE

14

Solution

Verify the eight trigonometric identities using the right triangle in the figure. First we need to calculate the length of the hypotenuse. By the Pythagorean Theorem,

hyp2 hyp2 hyp2 hyp

= = = =

22 + 32 4+9 13 ò13.

opp = 3

hyp q adj = 2

Before verifying the identities, let us write the six trigonometric ratios for the given right triangle: sin θ =

3 13

3 2 13 13 , tan θ = , cot θ = , csc θ = . , sec θ = 3 2 3 13 2 2

, cos θ =

Now we can verify the identities. tan θ ⋅ cot θ =

3 2 ⋅ =1 2 3

1 1 13 = = = csc θ 3 sin θ 3 13 1 1 13 = = = sec θ 2 cos θ 2 13

We know that tan θ =

3 13 and sec θ = , so by substitution, 2 2

3 13 2 ( )2 +1=( ) 2 2 9 13 +1= 4 4 13 13 = . Therefore, tan 2θ + 1 = sec 2θ . 4 4

We know that cot θ =

13 2 , so by substitution, and csc θ = 3 3

2 13 2 ( )2 +1=( ) 3 3 4 13 +1= 9 9 13 13 = . Therefore, cot2 θ + 1 = csc2 θ, 9 9 102

Algebra 10

sin θ = cos θ

3

2

13 3 cos θ = = tan θ , = 2 2 sin θ

13 2 = = cot θ, 3 3

13

13

sin 2 θ + cos 2 θ = (

3

) 2 +(

13

2 13

)2 =

9 4 13 + = =1. 13 13 13

Check Yourself 5 2 1. Verify the eight trigonometric identities for the acute angle θ in a right triangle if sin θ = . 5 2. Verify the eight trigonometric identities for a right triangle with sides of length 7, 24 and

25 units.

4 3. Let α be an acute angle in a right triangle such that sin α = . 5 Evaluate cos α ⋅ (tan α + cot α). π 21π cos 2 + sin 2 +1 5 5 4. Evaluate . π π 2 – tan ⋅ cot 7 7 Answers 5 3. 4. 2 4

2. Simplifying Trigonometric Expressions In the previous section we studied the eight most common trigonometric identities. These identities are useful when we are simplifying trigonometric expressions. Let us look at some examples. EXAMPLE

15

Solution

Simplify cos x ⋅ tan x. We can use the identity tan x = cos x ⋅ tan x = cos x ⋅

sin x cos x

= cos x ⋅

sin x cos x

sin x : cos x

(substitute) (cancel)

= sin x. So cos x ⋅ tan x = sin x. Trigonometry: Functions, Equations and Inequalities

103

EXAMPLE

16

a. Simplify tan x ⋅ cos x ⋅ csc x. b. Simplify cos3 x + sin2 x ⋅ cos x.

Solution a. We know tan x =

1 sin x . Hence, and csc x = sin x cos x

tan x ⋅ cos x ⋅ csc x = =

sin x 1 ⋅cos x ⋅ cos x sin x

(substitute)

sin x 1 ⋅ cos x ⋅ =1. sin x cos x

(cancel)

So tan x ⋅ cos x ⋅ csc x = 1. b. Since cos x is the common factor in both terms of the expression, let us factorize the expression: cos3 x + sin2 x ⋅ cos x = cos x ⋅ (cos2 x+ sin2 x) (factorize) = cos x ⋅ 1 = cos x

(using cos2 x + sin2 x = 1)

So cos3 x + sin2 x ⋅ cos x = cos x. EXAMPLE

17

Solution

Simplify

sec x – cos x . tan x

We know tan x =

sin x 1 and sec x = . Hence, cos x cos x

1 – cos x sec x – cos x cos x = sin x tan x cos x

104

(by substitution)

1 cos 2 x − = cos x cos x sin x cos x

(equalize the denominators)

1 − cos 2 x = cos x sin x cos x

(simplify the numerator)

1 − cos 2 x cos x = sin x cos x

(cancel the common divisor)

Algebra 10

=

sin 2 x sin x

(using sin2 x + cos2 x = 1)

= sin x. As a result,

EXAMPLE

18

Solution

(by cancellation)

sec x − cos x = sin x. tan x

2 x Simplify 2+ tan −1 2 sec x

We know tan x =

2+ tan 2 x − 1= sec 2 x

=

sin x 1 and sec x = . Hence, cos x cos x sin x 2 ) cos x – 1 1 2 ( ) cos x

2+(

(by substitution)

sin 2 x cos 2 x – 1 1 cos 2 x

2+

2 ⋅ cos 2 x sin 2 x + 2 cos 2 x − 1 = cos x 1 cos 2 x

(equalize the denominators in the numerator)

2 ⋅ cos 2 x + sin 2 x cos 2 x = −1 1 cos 2 x

(simplify the numerator)

2 ⋅ cos 2 x + sin 2 x =

cos 2 x 1

−1

(cancel the common divisor)

cos 2 x

= cos2 x + cos2 x + sin2 x – 1

(2 ⋅ cos2 x = cos2 x + cos2 x)

= cos2 x + 1 – 1

(using cos2 x + sin2 x = 1)

= cos2 x + 1 – 1

(by cancellation)

= cos2 x. 2 x As a result, 2+ tan − 1 = cos 2 x. 2 sec x

Trigonometry: Functions, Equations and Inequalities

105

Check Yourself 6 Simplify the expressions. 1. cos x ⋅ tan x

2.

1+ csc x cos x + cot x

3.

1 1 + 1 − sin x 1+ sin x

Answers 1. sin x

2. sec x

3. 2 ⋅ sec2 x

3. Verifying Trigonometric Identities In the previous section we learned how to write a trigonometric expression in an alternative (simpler) form using the eight basic identities. This means that we can derive other identities using the eight basic identities. In this section we will learn how to verify a given trigonometric identity. To verify an identity, we try to show that one side of the identity is the same as the other side. We take either the left-hand or right-hand side of the identity and do algebraic operations to obtain the other side. Generally, it is easier to begin working with the more complex side of the identity. Let us look at some examples. EXAMPLE

19

Solution

Verify the identity sin x ⋅ cot x = cos x. We can begin with either the left-hand side or the right-hand side. In this example we will show both approaches. Working on the left-hand side: sin x ⋅ cot x = sin x ⋅

= sin x ⋅

cos x sin x

cos x = cos x. sin x

(by substitution) (simplify)

We have obtained the right-hand side and the verification is complete. Working on the right-hand side: cos x = cos x ⋅ 1 = cos x ⋅ tan x ⋅ cot x sin x ⋅ cot x = cos x ⋅ cos x = cos x ⋅

sin x ⋅ cot x = sin x ⋅ cot x. cos x

(by substitution) (tan x ⋅ cot x = 1) (by substitution) (simplify)

We have obtained the left-hand side and the verification is complete. 106

Algebra 10

EXAMPLE

20

Solution

Verify the identity csc x = cos x ⋅ (tan x + cot x). Since the right-hand side is more complex than the left-hand side, let us try to transform the right-hand side into the left-hand side. We know tan x =

sin x cos x and cot x = , so cos x sin x

cos x ⋅ (tan x+ cot x) = cos x ⋅( = cos x ⋅ (

sin x cos x + ) cos x sin x sin 2 x + cos 2 x ) cos x ⋅ sin x

1 ⎛ ⎞ = cos x ⋅ ⎜ ⎟ sin ⋅ cos x x ⎝ ⎠ =

1 = csc x. sin x

We have obtained the other side of the identity and the verification is complete.

EXAMPLE

21

Solution

2 Verify the identity (sin x + cos x) = 2+ sec x ⋅ csc x. sin x ⋅ cos x

Let us begin with the left-hand side as it is more complex. (sin x + cos x) 2 sin 2 x+(2 ⋅ sin x ⋅cos x)+ cos 2 x = sin x ⋅ cos x sin x ⋅cos x

(expand the numerator)

=

(2 ⋅ sin x ⋅ cos x)+1 sin x ⋅ cos x

(cos2 x + sin2 x = 1)

=

2 ⋅ sin x ⋅ cos x 1 + sin x ⋅ cos x sin x ⋅ cos x

(separate the fractions)

= 2+ csc x ⋅ sec x

(simplify)

We have obtained the right-hand side of the identity and the verification is complete. Trigonometry: Functions, Equations and Inequalities

107

EXAMPLE

22

Solution

Verify the identity

tan x = sec x − cos x. csc x

Let us work on the left-hand side. sin x tan x cos x = 1 csc x sin x

(by substitution)

=

sin x sin x ⋅ cos x 1

=

sin 2 x cos x

=

1 − cos 2 x cos x

(cos2 x + sin2 x = 1)

=

1 cos 2 x − cos x cos x

(separate the fractions)

(invert the denominator and multiply) (multiply)

= sec x − cos x

(substitute and simplify)

We have obtained the right-hand side of the identity and so the verification is complete. EXAMPLE

23

Solution

Verify the identity

cos x 1 + sin x = . 1 − sin x cos x

Begin with the left-hand side. cos x cos x 1+ sin x ⋅ = 1 − sin x 1 − sin x 1+ sin x

(multiply by 1)

=

cos x ⋅ (1+ sin x) 1 − sin 2 x

(write the product)

=

cos x ⋅ (1+ sin x) cos 2 x

(cos2 x + sin2 x =1)

1+ sin x (cancel the common factor) cos x This is the right-hand side of the identity, so the verification is complete. =

Check Yourself 7 Verify the identities. 1. sec x – cos x = sin x ⋅ tan x 3. 108

1+ sin x cos x + = 2 sec x cos x 1+ sin x

2.

cos x = csc x − sin x sec x ⋅ sin x

Algebra 10

4. Cofunctions We have studied the trigonometric functions of certain angles and the trigonmetric ratios between the sides and angles of a right triangle. In this section we will look at the relation between the trigonometric ratios of complementary angles. πR . Consider the right 2 triangle ABC with acute angles θ and α shown in the figure. θ and α are complementary angles. b C We can also write sin θ = cos α = . In other a words, the sine of θ and the cosine of its a

Recall that complementary angles are angles whose sum is 90°, i.e.

complement are equal. We say that sine and cosine are cofunctions. Looking at the b q triangle we can also write tan θ = cot α = B c (so tangent and cotangent are cofunctions) a and sec θ = csc α = (i.e. secant and cosecant are cofunctions). c In other words, for θ + α = 90°, θ = 90° – α we have

a

b

c

A

sin θ = sin(90° – α) = cos α tan θ = tan(90° – α) = cot α sec θ = sec(90° – α) = csc α. For example,

sin 43° = cos(90° – 43°) = cos 47°, cos 26° = sin 63°, tan 3° = cot 87°, sec 18° = csc 72°, π π π 3π sin = cos( – ) = cos and 5 2 5 10 cos

EXAMPLE

24

3π π 3π π = sin( – ) = sin , etc. 8 2 8 8

Evaluate each expression. a. tan 1° ⋅ tan 2° ⋅ tan 3° ⋅ ... ⋅ tan 88° ⋅ tan 89° 2 b. sin

π 7π π 5π +[tan ⋅ tan ]+ sin 2 7 18 9 14

Trigonometry: Functions, Equations and Inequalities

109

Solution

a. The angles in each pair (89°, 1°), (88°, 2°), ..., (46°, 44°) are complementary. Because tangent and cotangent are cofunctions, tan 89° = cot 1°, tan 88° = cot 2°, ... , tan 46° = cot 44°. So tan 1° ⋅ tan 2° ⋅ tan 3° ⋅ ... ⋅ tan 88° ⋅ tan 89° = tan 1° ⋅ tan 2° ⋅ tan 3° ⋅ ... ⋅ tan 44° ⋅ tan 45° ⋅ cot 44° ⋅ ... ⋅ cot 2° ⋅ cot 1° = 1 ⋅ 1 ⋅ 1 ⋅ ... ⋅ tan 45° ⋅ ... ⋅ 1 ⋅ 1 = 1. π 5π π 5π π since + = . b. The complement of is 7 14 7 14 2 Similarly, the complement of 7π π 7π π π is since + = . 18 9 18 9 2

So sin

5π π 7π π = cos and tan = cot since these are cofunctions. 14 7 18 9

So sin 2

π 7π π 5π π π π π + tan ⋅ tan + sin 2 = sin 2 +[cot ⋅tan ]+cos 2 7 18 9 14 7 9 9 7 π π π π = sin 2 + cos 2 + cot ⋅ tan 7 7 9 9  

   1

1

=1+1 = 2.

Check Yourself 8 1. Write the cofunction of each function. a. tan 15° d. sin

b. cos 36°

π 12

e. cot

c. sec 77°

2π 5

f. tan

2π 7

2. Evaluate each expression. a. tan 5° ⋅ tan 25° ⋅ tan 45° ⋅ tan 65° ⋅ tan 85° π 5π b. tan ⋅ tan – cos 2 27° – cos 2 63° 7 14

Answers 1. a. cot 75° 2. a. 1 b. 0 110

b. sin 54°

c. csc 13°

d. cos

5π 12

e. tan

π 10

f. cot

3π 14 Algebra 10

Eratosthenes was a famous mathematician and the head of the famous library in Alexandria, Egypt. In 240 BC he calculated the Earth's circumference using trigonometry and his knowledge of the angle of elevation of the Sun at the summer solstice in the Egyptian cities of Alexandria and Syene (now called Aswan). Eratosthenes’ calculation was based on the assumptions that the Earth is a sphere and that the sun is so far away that we can consider its rays to be parallel. Eratosthenes

compared

observations

made

in

Alexandria,

where

the

noontime Sun at the summer solstice was 7° away from straight overhead (the zenith), to observations in Syene in southern Egypt, where the Sun was exactly at its zenith. The distance between the cities was known to be about 5000 stadia, roughly equal to 800 km (the stadion, plural stadia, was an old unit of measurement such that 1 stadia ≅ 160 m). Therefore, Eratosthenes calculated the entire 360° circle of the Earth to be (360/7) ⋅ 5000 stadia, which is about 260,000 stadia, or 41,000 km. a2

B

parallel rays from the Sun

a1

local horizon

A a 1– a2

a1 = altitude of Sun at site 1 a2 = altitude of Sun at site 2

circumference of the Earth = arc length AïB ⋅ (360° / (α – α1))

EXERCISES

3 .1

A. Angles and the Unit Circle 1. In which quadrant does each angle lie? a. 275° d. –530°

b. 228° 28π e. 3

5. Find the primary directed angle of 1720 grads in degrees.

c. 185° 125 π f. 6

6. Solve for x if x ∈ \, k ∈ ]. a. 2x – 120° = 90° + (k ⋅ 360°)

2. Find the value of the real number which corresponds to each angle on the unit circle. Give your answer to three decimal places. a. 45° 3π d. 7

b. 105° 26π e. 5

c. –70°

b. x + π = π + x +( k ⋅ 2 π) 3 4 3 4 c. 4x = k ⋅ 360° d. 3x − 150 ° = π +( k ⋅ 2 π) 6

C. Trigonometric Ratios B. Primary Directed Angles 3. Find the primary directed angle of each angle, using the same unit. a. 1234° c. 190 π 9

b. –4321° 90 π d. − 19

1 is given. Evaluate each ratio, given 17 that x is in the first quadrant.

7. tan x =

a. cot x

b. sin x

c. cos x

8. In the figure, m(∠A) = 90°, 4. An arc lies in standard position on the unit circle. Find the coordinates of the terminal point of the arc if the arc has length a. π.

112

b. 5π . 4

C

AD = 3, DB = 5 and

x

m(∠ADC) = 60°. Calculate BC = x.

60° B

5

D

3

A

Algebra 10

9. In the figure,

13. The figure shows a

D

m(∠DAC) = 45°,

C

m(∠BAC) = 30° and BC = 3ñ2. 45°

30°

A

D

V

F

E

right triangles, increasing in size from right to left. The length of the hypotenuse of the eighth triangle in the series is 72 units. Calculate AO.

3ñ2

Calculate DC = x.

10.

›› series of 30°-60°-90°

x

B

IV D

III

C II B

O

30°

C

30° 120°

3 A

B

8

In the figure, DC || AB, m(∠D) = 30°, m(∠B) = 120°, AD = 3 and AB = 8. Calculate DC.

D. Trigonometric Identities 14. Simplify the expressions. a. csc x ⋅ tan x

11. In the figure,

A

m(∠C) = 120°, m(∠A) = 30° and AB = 26. Calculate the height h.

c. tan x + cot x

30° 26

e. tan x sec x

h

b. sec2 x – tan2 x 1 + sin x d. 1 + csc x f.

cot x − 1 1 − tan x

120° B

C

12. Calculate the length x in each figure. ›

a.

15. Simplify the expressions.

A M

N

›

60° 4

b.

x B

K

L

b.

A 3 15° H x

Trigonometry: Functions, Equations and Inequalities

cos x sec x + tan x

c. sin4 x – cos4 x + cos2 x

C

B

a. (sec x – tan x)2 (1 + sin x)

C

d.

sin x − csc x 1 − cos x

e.

1 1 + 2 sec x csc2 x

f. (tan x + sin x)2 + (1 + cos x)2 113

I A

16. Verify the identities.

18. Evaluate the expressions.

a. (1 – cos x)(1 – cos x) = sin2 x b. cos x + sin x =1 sec x csc x c.

(sin x + cos x) 2 sin 2 x − cos 2 x = 2 2 sin x − cos x (sin x − cos x) 2

d.

1 − sin x = (sec x − tan x) 2 1+ sin x

tan

a.

b.

2π 3π π ⋅ tan – sin 2 7 14 10 2π 1 – cos 2 5

tan 25° ⋅ tan65° +2 sin2 25° + sin 2 65°

e. csc x – sin x = cos x cot x f. (cot x – csc x)(cos x + 1) = –sin x

19. Evaluate sin 2

π 3π π 5π + sin 2 + cos 2 + cos 2 . 8 8 12 12

17. Verify the identities. a. 1+ tan x = cos x+ sin x 1 − tan x cos x − sin x b.

cos x sin x − cos x = 1 − sin x cos x − cot x

c.

1 1 − = 2 sec x tan x 1 − sin x 1 + sin x

d.

tan x − cot x = sin x cos x tan 2 x − cot 2 x

e. 1+ sin x = (tan x + sec x )2 1 − sin x f.

114

1+ cos x 2 = (cot x + csc x ) 1 − cos x Algebra 10

In the previous section we defined the trigonometric ratios in terms of the side lengths of a right triangle. In this section we will study the same ratios as ratios on the unit circle and as functions. To define a trigonometric function we must find its domain, range, graph and inverse. Recall that the domain of a function f(x) is the largest set of real x-values for which f(x) is defined. For example, the domain of a polynomial function is . The range of function is a set which includes at least all images of the elements in its domain. The largest possible range of a real function is . Recall also the definition of increasing and decreasing function: we say that f(x) is an increasing function if f(x1) < f(x2) when x1 < x2, and f(x) is a decreasing function if f(x1) > f(x2) when x1 < x2. If f(x1) = f(x2) for every x1 < x2 then f(x) is a constant function. With these basic definitions we are ready to look at the properties of trigonometric functions.

A. TRIGONOMETRIC FUNCTIONS AND THEIR PROPERTIES 1. The Sine Function Consider the unit circle and an angle α in standard position on the circle, as shown at the right. In the figure, m(∠AOB) = α and OB = 1. Let us draw perpendiculars from point B to the

B

D 1 a O

A C

x- and y-axis, as shown. BC BC = = BC = OD. This OB 1 is true for any angle α. So we can say that the

We know sin α =

sine axis

y-axis is the sine axis. By the definition of the unit circle, we can also say –1 ≤ sin α ≤ 1 and the angle α can be any real number. We can define a function from the real numbers to the interval [–1, 1] with the rule f(x) = sin x. This function is called the sine function. This function has the following properties: The sine function corresponds to the y-coordinates of points on the unit circle.

1. Domain =

.

2. Range = [–1, 1] 3. sin x > 0 in the interval (0, π).

Trigonometry: Functions, Equations and Inequalities

115

4. sin x < 0 in the interval (π, 2π). 5. sin x is increasing in the first and fourth quadrants. 6. sin x is decreasing in the second and third quadrants. 7. sin x is a continuous function. By using the unit circle we can form the following table:

EXAMPLE

25

Solution

degrees

0°

radians

0

sin α

0

90° π 2 1

180° π 0

270° 3π 2 –1

360° 2π 0

Calculate sin 0° + 3 sin 90° – 2 sin 180° - sin 270°. We can find these values from the unit circle: sin 0° + 3 sin 90° – 2 sin 180° – sin 270° = 0 + (3 ⋅ 1) – (2 ⋅ 0) – (–1) = 3 + 1 = 4.

EXAMPLE

26

Solution

Find the domain and range of f(x) = 3 sin x + 2. We know that the domain of sin x is

.

The sine function is a continuous function, so we can find the range by calculating the maximum and minimum values of f(x). The maximum value of sin x is 1, so the maximum value of f(x) is 3 ⋅ 1 + 2 = 5. The minimum value of sin x is –1, so the minimum value of f(x) is 3 ⋅ (–1) + 2 = –1. In conclusion, the range of f(x) is [–1, 5].

2. The Cosine Function Look at the unit circle in the figure. OC OC = = OC, and this is OB 1 true for any angle α. So we can say that the

We know cos α =

B 1

x-axis is the cosine axis. From the unit circle we have –1 ≤ cos α ≤ 1 and the angle α can be any real number. The cosine function corresponds to the x-coordinates of points on the unit circle.

116

a O

A C

cosine axis

We can define a function from the real numbers to the interval [–1, 1] with the rule f(x) = cos x. This function is called the cosine function. It has the following properties: Algebra 10

1. Domain = 2. Range = [–1, 1]

π π 3. cos x > 0 in the interval (– , ). 2 2 π 3π 4. cos x < 0 in the interval ( , ). 2 2 5. cos x is increasing in the third and fourth quadrants.

6. cos x is decreasing in the first and second quadrants. 7. cos x is a continuous function. By using the unit circle we can form the following table:

EXAMPLE

27

Solution

degrees

0°

radians

0

cos x

1

90° π 2 0

180° π –1

270° 3π 2 0

360° 2π 1

Calculate cos 360° – 2 cos 90° – 3 cos 180° + cos 270° – cos 0°. We can use the values in the table we have just seen: cos 360° – 2 cos 90° – 3 cos 180° + cos 270° – cos 0° = 1 – (2 ⋅ 0) – [3 ⋅ (–1)] + 0 – 1 = 1 + 3 – 1 = 3.

EXAMPLE

28

Solution

Find the domain and range of f(x) = 4 cos x – 2. We know that the domain of cos x is

. So the domain of f(x) is also

.

The maximum value of cos x is 1, so the maximum value of f(x) is 4 ⋅ 1 – 2 = 2. The minimum value of cos x is –1, so the minimum value of f(x) is 4 ⋅ (–1) – 2 = –6. In conclusion, the range of f(x) is [–6, 2].

EXAMPLE

29

Solution

Find the range of y = sin 3x ⋅ cos 2x + cos 3x ⋅ sin 2x + 1. We know (sin 3x ⋅ cos 2x) + (cos 3x ⋅ sin 2x) = sin(3x + 2x) = sin 5x by the sum and difference formulas. So the equation simplifies to y = sin 5x + 1. Also, –1 ≤ sin 5x ≤ 1, so (–1 + 1) ≤ (sin 5x + 1) ≤ (1 + 1). In conclusion, the range of y is 0 ≤ y ≤ 2.

Trigonometry: Functions, Equations and Inequalities

117

EXAMPLE

30

Solution

Calculate sin π + cos

13π 11π + sin17 π – 2 ⋅cos13 π+ sin . 2 2

We can find the answer by using coterminal angles and quadrantal angles for each term. sin π = 0 cos

13π π π = cos( + 3 ⋅ 2 π) = cos = 0 2 2 2

sin17 π = sin( π+8 ⋅ 2 π) = sin π = 0 cos13 π = cos( π+6 ⋅ 2 π) = cos π= –1 sin

11π 3π 3π = sin( + 2 ⋅ 2 π) = sin = –1 2 2 2

Combining these results give us sin π + cos

13π 11π + sin17 π – 2 ⋅cos13 π+ sin = 0+0+0 – 2(–1)+(–1 ) =1. 2 2

3. The Tangent Function The figure opposite shows the unit circle and an angle α. Let us draw a line through the point A(1,0) perpendicular to the x-axis. Then we extend OB such that OB and the new line intersect at point D. Then BC DA DA = = = DA. We can do this 1 OC OA π 3π 5π for all values of x except ± , ± , ± ... . 2 2 2 We can call this new line the tangent axis.

B

A

a O

tangent axis D

C

tan α =

The tangent function corresponds to the y-coordinates of points on the tangent axis.

We can see in the figure that the tangent axis has infinite length. So we can define a π function from – { + kπ}, k ∈ to such that f(x) = tan x. This function is called the 2 tangent function. It has the following properties: π 1. Domain = – { + kπ}, k ∈ 2 2. Range = π 3π ), ( π, ) 2 2 π 3π 4. tan x < 0 on ( , π), ( , 2 π) 2 2

3. tan x > 0 on (0, Can you see why this function is called the tangent function?

5. tan x is always increasing. 6. tan x =

118

sin x , cos x ≠ 0. cos x Algebra 10

By using the unit circle we can form the following table:

EXAMPLE

31

Solution

degrees

0°

radians

0

tan x

0

90° π 2 undefined

180° π 0

270° 3π 2 undefined

360° 2π 0

Find the domain and the range of f(x) = tan 3x + 4. π π kπ + kπ gives us x ≠ + , k ∈ . 2 6 3 π kπ So the domain is {x | x ∈ , x ≠ + , k ∈ }. 6 3

Domain: 3x ≠

Range: tan 3x ∈

, so the range is (–∞, ∞).

4. The Cotangent Function

The cotangent function corresponds to the x-coordinates of points on the cotangent axis.

The figure shows the unit circle and a line through the point B(0, 1) parallel to the x-axis. If we extend OC such that OC and the new line intersect at point D, then OF OE OE cot α = = = = OE = BD. CF DE 1 We can do this for all values of α except 0, ±π, ±2π... . We can call this new line the cotangent axis.

D

B

cotangent axis

C a O

F A

E

We can see in the figure that this axis has infinite length. So we can define a function from – {kπ}, k ∈ to such that f(x) = cot x. This function is called the cotangent function. It has the following properties: 1. Domain =

– {kπ}, k ∈

2. Range = π 3π ), ( π, ). 2 2 π 3π 4. cot x < 0 on ( , π), ( , 2 π) 2 2

3. cot x > 0 on (0,

5. cot x is always decreasing. 6. cot x =

cos x , sin x ≠ 0 sin x

Trigonometry: Functions, Equations and Inequalities

119

By using the unit circle we can form the following table: degrees

0°

radians

0

cot x

undefined

90° π 2 0

180° π undefined

270° 3π 2 0

360° 2π undefined

5. The Secant Function The figure shows the angle ∠POC = α and the unit circle. Let us draw a tangent line to the unit circle at point P and say that the

P

intersection point of the tangent line and the OC OC = = OC. x-axis is C. Then sec α = 1 OP

a O

C A

We can do this for all values of α except π 3π 5π ± , ± , ± , ... because at these points 2 2 2 the tangent line will be parallel to the x-axis. We can see in the figure that as α approaches 90°, sec α approaches ∞. π – { + kπ}, k ∈ to – (–1, 1) such that f(x) = sec x. This 2 function is called the secant function. It has the following properties:

We can define a function from

1. Domain = The secant function corresponds to the x-coordinates of points on the x-axis with x ∉ (–1, 1).

2. Range =

π – { + kπ}, k ∈ 2

– (–1, 1)

π π 3. sec x > 0 on (– , ) 2 2 π 3π 4. sec x < 0 on ( , ) 2 2 1 , cos x ≠ 0. cos x By using the unit circle we can form the following table:

5. sec x =

120

degrees

0°

radians

0

sec x

1

90° π 2 undefined

180° π –1

270° 3π 2 undefined

360° 2π 1 Algebra 10

6. The Cosecant Function The figure shows the angle ∠POA = α and D

the unit circle. Let us draw the tangent to

x P

the unit circle at point P and say that the intersection point of the tangent line and the OD OD = = OD. y-axis is D. Then csc α = OP 1 We can do this for all values of x except 0,

a O

A

±π, ±2π, ±3π, ... because at these points the tangent line will be parallel to the y-axis. We see in the figure that as α approaches 0°, csc α approaches ∞. So we can define a function from

– {kπ}, k ∈

to

– (–1, 1) such that f(x) = csc x. This

function is called the cosecant function. It has the following properties: – {kπ}, k ∈

1. Domain = The cosecant function corresponds to the y-coordinates of points on the y-axis with y ∉ (–1, 1).

2. Range =

– (–1, 1)

3. csc x > 0 on (0, π) 4. csc x < 0 on (π, 2π) 5. csc x =

1 , sin x ≠ 0. sin x

By using the unit circle we can form the following table: degrees

0°

radians

0

csc x

undefined

90° π 2 1

180° π undefined

270° 3π 2 –1

360° 2π undefined

7. Mixed Examples EXAMPLE

32

Solution

π Given the function f(x) = 2sin x + cos 2x, find f ( ) . 6 π π π 1 1 3 f ( ) = 2 ⋅ sin + cos(2 ⋅ ) = 2 ⋅ + = . 6 6 6 2 2 2

Trigonometry: Functions, Equations and Inequalities

121

EXAMPLE

33

Solution

Calculate (sin

π 3π 3π ⋅ cos π) – (tan π ⋅ cot )+ sec 2 π– csc . 2 2 2

π 3π We know that sin = 1, cos π = –1, tan π = 0, cot = 0, sec 2π = 1 and 2 2 3π csc = –1. So 2 (sin

π 3π 3π ⋅ cos π) – (tan π ⋅ cot )+ sec 2 π– csc =1 ⋅(–1) – 0 ⋅0+1 – (– 1) =1. 2 2 2

Note

p 2

We can summarize the signs of the functions cos x, sin x, tan x and cot x as shown in the figure.

cos x tan x

O p

EXAMPLE

34

Solution

sin x cot x

2p

3p 2

Find the sign of each ratio. sin 200°, cos 122°, tan 312°, cot 300°, sec 120°, csc 77° We can use the figure above. 200° is in the third quadrant, so sin 200° is negative. 122° is in the second quadrant, so cos 122° is negative. 312° is in the fourth quadrant, so tan 312° is negative. 300° is in the fourth quadrant, so cot 300° is negative. 120° is in the second quadrant, so sec 120° is negative. 77° is in the first quadrant, so csc 77° is positive.

EXAMPLE

35

Solution

Write a = sin 65°, b = cos 124° and c = tan 55° in ascending order. Let us draw the given angles on the unit circle. We know that tangent function is an increasing function, so if 55° > 45° then tan 55° > tan 45° = 1. So c > 1. Also from the figure we have a = sin 65° is positive and smaller than 1, and b = cos 124° is negative. So b < a < c.

122

65°

124°

55° sin 65°

cos 124°

cos x

tan 55°

Algebra 10

EXAMPLE

36

Solution

Write a = sin 50°, b = cos 50°, c = tan 50° and d = sec 50° in ascending order. Look at the figure. At 45°, sin 45° = cos 45° and cos 45° < tan 45° = 1. Also, if 90° > x > 45° then sin x > cos x and tan x > 1. So we can write cos 50° < sin 50° < tan 50°. What about sec 50°? sin50 ° 1 tan50 ° = and sec50 °= . cos50 ° cos50 °

C tan 50°

B sin 50°

A

O cos 50°

D

We know sin 50° < 1, so tan 50° < sec 50°. So b < a < c < d. EXAMPLE

37

Solution

Find the domain and range of y =2 sin x + 3 tan 2x. Let us find the domain and range of 2 sin x and 3 tan 2x separately: f(x) = 2 sin x: domain =

In this type of question, begin by finding the domain and range of the individual functions. Then the domain is the intersection of the domains and the range is the union of the ranges.

EXAMPLE

38

Solution

and range = [–2, 2]. π kπ g(x) = 3 tan 2x: domain = − { + }, k ∈ and range = 4 2 So the domain of the function y is The range of y is

∪ [–2, 2] =

∩

π kπ − { + }, k ∈ 4 2

=

.

π kπ − { + }, k ∈ 4 2

.

.

Find the domain and the range of y = cos2 x + cos x. The domain of both cos2 x and cos x is

. So we can say that the domain of y is

∩

=

.

To find the range, we can use the fact that cos2 x and cos x are continuous functions. The range is therefore the interval between the minimum and maximum points of y. y = ax2 + bx + c is the equation of a parabola. The vertex of the parabola is V(r, k) where b and r=– 2a k =y(k).

Let us use the substitution t = cos x, then we have y = t2 + t. 1 1 b So y is a parabola and its vertex is V(r, k) where r = – and =– =– 2a 2 ⋅1 2 1 1 1 1 1 1 k = y(– ) = (– )2 – = – = – . 2 2 2 4 2 4 1 So the minimum value of y is – . 4 If we substitute the maximum value of cos x in the function we will get the maximum value of y. The maximum value of cos x is 1, so the maximum value of y is 12 + 1 = 2. 1 In conclusion, the range of y is [– , 2]. 4

Trigonometry: Functions, Equations and Inequalities

123

Check Yourself 9 1. Evaluate each expression. a. sin 90° + cos 180° – 2 ⋅ tan 180° – cot 90° b. 3 ⋅ cos 0º + 3 ⋅ tan 0º – 2 ⋅ cot 270° + sin 270° 3π π 3π + 2 ⋅ cos – cot 2 2 2 7π 21π e. sin 3π + cos + tan17 π+ cot 2 2

c. sin

π 13π – cot 2 2 3π 5π sec17 π + csc – tan6 π+ cot 2 2

d. sec 3π + 2 ⋅ csc f.

2. Find the domain of each function. a. f(x) = 3 sin(5x + 2)

b. y = 4 cos 5x

c. y = 4 tan 3x – 2

d. f(x) = cot(3x – 5) + 2

e. f(x) = 3 csc 2x – 2

f.

g. y = 2 sin 3x + cos 2x

h. y = 2 tan 3x – 3 cot 5x

π y = sec(2 x + ) 3

3. Find the range of each function. a. y = 3 sin 2x + 1

b. y = 5 sin (4x + 1) – 3

c. f(x) = 3 sin 7x + 4

d. y = 2 cos 5x – 4

e. y = 13 cos (2x – 3)

f. f(x) = 5 cos x + 3

g. y = tan 3x – 1

h. f(x) = 2 tan (4x – 2) + 6

i. f(x) = 2 sec 3x

j. y = 4 csc 7x – 3

k. y = 2 sec (x + 2)

l. y = 2 tan x + cot 2x

4. Find the sign of each ratio. a. sin 233°

b. cos 129°

c. tan 448°

d. cot 322°

e. tan 198°

f. sec 121°

g. csc 167°

h. sin

7π 5

l.

3π 4

i.

cos

17 π 7

j.

tan

6π 5

k. cot

13π 19

sec

5. Write each set of ratios in ascending order. a. x = sin 45°, y = cos 74° and z = tan 155° b. a = sin 130°, b = cos 130°, c = tan 130° and d = sec 130° c. m = sin 119°, n = tan 244° and r = cot(–12°) Answers 1. a. 0 b. 2 c. –1 d. 1 e. 0 f. –2 2. a. f. 124

b.

c.

π kπ − { + }, k ∈ 6 3

π kπ − { + }, k ∈ 12 2

g.

h.

d.

5 kπ − { + }, k ∈ 3 3

e.

−{

kπ }, k ∈ 2

π kπ nπ −{ + ∪ }, k, n ∈ 6 3 5 Algebra 10

3. a. [–2, 4] b. [–8, 2] c. [1, 7] d. [–6, –2] e. [–13, 13] f. [–2, 8] g.

h.

i. (–∞, –2] ∪ [2, ∞) j. (–∞, –7] ∪ [1, ∞) k. (–∞, –2] ∪ [2, ∞) l. 4. a. – b. – c. + d. – e. + f. – g. + h. – i. + j. + k. – l. – 5. a. z < y < x b. d < c < b < a c. r < m < n

B. CALCULATING TRIGONOMETRIC VALUES 1. Trigonometric Values of Quadrantal Angles quadrantal angle

Definition

If the terminal side of an angle coincides with a coordinate axis then the angle is called a quarantal angle. If an angle is not quadrantal, it is called a nonquadrantal angle.

y B(0, 1) p 90° 2 A¢(–1, 0) 180° p

0° A(1, 0)

x

360° 2p

O

3p 2 270° B¢(0, –1) the angles 0+ kp and p + kp (k Î ¢) 2 are quadrantal angles

We can calculate the trigonometric values of quadrantal angles by observing at the points at which the terminal sides of the angles intersect the unit circle. Trigonometric Values of Quadrantal Angles

EXAMPLE

39

Solution

q in degrees

q in radians

sin q (y)

cos q (x)

tan q (y/x)

cot q (x/y)

sec q (1/x)

csc q (1/y)

0°

0

0

1

0

undefined

1

undefined

90°

p 2

1

0

undefined

0

undefined

1

180°

p

0

–1

0

undefined

–1

undefined

270°

3p 2

–1

0

undefined

0

undefined

–1

360°

2p

0

1

0

undefined

1

undefined

Evaluate each expression. a. sin 0° + cos 270° + tan 180° – cot 90°

b. sin 90° – tan 180° + cot 270° – cos 180°

We can use the table above. a. 0 + 0 + 0 – 0 = 0

Trigonometry: Functions, Equations and Inequalities

b. 1 – 0 + 0 – (–1) = 2 125

EXAMPLE

40

Solution

Evaluate each expression. π 3π a. sin π + cos + tan 0 − cot 2 2 a. 0 + 0 + 0 – 0 = 0

b. sin

3π π − tan π − cot − cos 0 2 2

b. –1 – 0 – 0 – 1 = –2

EXAMPLE

41

Evaluate sin 1710° – cos 2520° + cot 450° – tan 900°.

Solution The angles are all greater than 360° so we begin by finding the primary directed angle of each term. sin 1710° = sin (4 ⋅ 360° + 270°) = sin 270° cos 2520° = cos (7 ⋅ 360° + 0°) = cos 0° cot 450° = cot (360° + 90°) = cot 90° tan 900° = tan (2 ⋅ 360° + 180°) = tan 180° Hence, sin 1710° – cos 2520° + cot 450° – tan 900° = sin 270° – cos 0° + cot 90° – tan 180° = –1 – 1 + 0 – 0 = –2.

Check Yourself 10 Evaluate each expression by using the table of trigonometric values for quadrantal angles. 1. 2 ⋅ sin 180° + tan π + 5 ⋅ cot 270° + 3 ⋅ cos 180°. π 4 ⋅ cos 0 − 10 ⋅ sin 2 2. 2 ⋅ sin 270 ° − cos π Answers 1. –3

2. 6

2. Reference Angle In this section we will learn how to find the trigonometric ratios of any angle in terms of the trigonometric ratios of a corresponding acute angle. We will use the special trigonometric ratios for 30°, 45° and 60° angles which we studied in section 1.2. Definition

reference angle The positive acute angle α which is formed by the terminal side of a nonquadrantal angle θ and the x-axis is called the reference angle for θ.

126

Algebra 10

Look at the figures. They show how to find the reference angle α for an angle θ in each quadrant. y

y

a=q O

x

O

If 0° < q < 90° then a = q.

If 90° < q < 180° then a = 180° – q.

y

y q

q a

q

a

x

x

O

If 180° < q < 270° then a = q – 180°.

O

a

x

If 270° < q < 360° then a = 360° – q.

The trigonometric reduction formulas help us to ‘reduce’ a trigonometric ratio to a ratio of an acute angle. If the acute angle is a common angle, this technique helps us to find the ratio. For example, imagine you need to find cot 300°. We can say that 300° = 270° + 30°. 3 . 3 To derive the reduction formulas, first we need to know the signs of the trigonometric

By the reduction formula for the cotangent, cot 300° = –tan 30° = − functions in each quadrant: 1.

third quadrant

π⎞ ⎛ ⎜ 0, ⎟ 2⎠ ⎝ ⎛π ⎞ ⎜ , π⎟ ⎝2 ⎠ 3π ⎞ ⎛ ⎜ π, ⎟ 2 ⎠ ⎝

fourth quadrant

⎛ 3π ⎞ ⎜ , 2π ⎟ ⎝ 2 ⎠

first quadrant second quadrant

Trigonometry: Functions, Equations and Inequalities

sin x

cos x

tan x

cot x

+

+

+

+

+

–

–

–

–

–

+

+

–

+

–

– 127

π 3π or in the reduction formula, the formula changes sine to cosine and 2 2 tangent to cotangent. If we have π or 2π in the formula, the function does not change.

2. If we have

3. Now we can combine these two pieces of information to get the reduction formulas: π π π sin( − x) =cos x, cos( − x) =sin x, tan( − x) =cot x 2 2 2 π π π sin( + x) =cos x, cos( + x) = − sin x, tan( + x) = −cot x 2 2 2 3π 3π 3π sin( − x) = − cos x, cos( − x) = −sin x, tan( − x) = cot x 2 2 2 3π 3π 3π sin( + x) = − cos x, cos( + x) =sin x, tan( + x) = −cot x 2 2 2 sin( π − x) =sin x, cos( π − x) = − cos x, tan( π − x) = −tan x sin( π + x) = − sin x, cos( π + x) = −cos x, tan( π + x) = tan x sin( − x) = − sin x, cos( − x) =cos x, tan( − x) = −tan x

EXAMPLE

42

Solution

π . 2 3π c. tan( + x) d. sin(2π – x) e. sin(π + x) f. cos(2π + x) 2

Simplify each expression, given that 0 < x < π a. sin( + x) 2

π b. cos( + x) 2

a. ( π + x) is in the second quadrant, so sin( π+ x) = cos x. 2 2 π π b. ( + x) is in the second quadrant, so cos( + x) = – sin x. 2 2 3 π c. ( + x) is in the fourth quadrant, so tan( 3 π+ x) = – cot x. 2 2 d. (2π – x) is in the fourth quadrant, so cot(2π – x) = –cot x. e. (π + x) is in the third quadrant, so sin(π + x) = –sin x. f. (2π + x) is in the first quadrant, so cos(2π + x) = cos x.

EXAMPLE

43

Solution

Simplify

cos(90º+x)+ sin(270º – x)+ sin(180º – x) . cos(– x) – cos(360º – x)+ sin(90º+ x)

Let us simplify each term using the reduction formulas: cos(90º + x) = –sin x, sin(270º – x) = –cos x, sin(180º – x) = sin x, cos(–x) = cos x, cos(360º – x) = cos x and sin(90º + x) = cos x. So cos(90º + x)+ sin(270º – x)+ sin(180º – x) – sin x– cos x+ sin x − cos x = = –1. = cos(– x) – cos(360º – x)+ sin(90º + x) cos x– cos x+ cos x cos x

128

Algebra 10

EXAMPLE

44

Solution

Find the reference angle for each angle. a. 30°

b. 150°

c. 215°

d. 317°

a. Since 0 < 30° < 90°, the reference angle for 30° is 30°. b. Since 90° < 150° < 180°, the reference angle for 150° is 180° – 150° = 30°. c. Since 180° < 215° < 270°, the reference angle for 270° is 270° – 215° = 55°. d. Since 270° < 137° < 360°, the reference angle for 317° is 360° – 317° = 43°.

Now that we can calculate reference angles we are ready to calculate the trigonometric value of a nonquadrantal angle. To do this, follow the steps: 1. Find the primary directed angle of the nonquadrantal angle and determine its quadrant. 2. Determine the sign of the function in this quadrant. 3. Calculate the reference angle for the given angle. 4. Find the trigonometric value of the reference angle and use it with the sign from step 2.

EXAMPLE

45

Solution

Find each trigonometric value by using a reference angle. a. cos 135°

b. sin 330°

c. sec 240°

d. csc 120°

e. sin 510°

f. cos 945°

g. tan (–930°)

h. cot (–675°)

a. 1. 135° is already a primary directed angle and it is in the second quadrant. 2. In the second quadrant, the cosine function is negative.

y sin a cos a tana cot a sec a csc a

(+) (–) (–) (–) (–) (+)

sin a cos a tana cot a sec a csc a

sin a cos a tana cot a sec a csc a

O (–) (–) (+) (+) (–) (–)

(+) (+) (+) (+) (+) (+) x

sin a cos a tana cot a sec a csc a

(–) (+) (–) (–) (+) (–)

3. The reference angle is α = 180° – 135° = 45°. 1 4. cos 135 ° = − cos 45° = − 2 b. 1. 330° is already a primary directed angle and it is in the fourth quadrant. 2. In the fourth quadrant, the sine function is negative. 3. The reference angle is α = 360° – 330° = 30°. 1 2 c. 1. 240° is a primary directed angle, third quadrant

4. sin 330 ° = − sin 30° = −

2. In the third quadrant, the secant function is negative. 3. The reference angle is α = 240° – 180° = 60°. 4. sec 240° = – sec 60° = –2

Trigonometry: Functions, Equations and Inequalities

129

d. 1. 120°: primary directed angle, second quadrant 2. In the second quadrant, the cosecant function is positive. 3. α = 180° – 120° = 60° 4. csc 120° = csc 60 ° =

2 3

e. 1. 510° = 360° + 150°. So the primary directed angle of 510° is 150° and it is in the second quadrant. 2. In the second quadrant, the sine function is positive. 3. α = 180° – 150° = 30° 4. sin 510 ° = sin 1 50° = sin 30 ° =

1 2

f. 1. 945° = (2 ⋅ 360°) + 225°. So the primary directed angle is 225° and it is in the third quadrant. 2. In the third quadrant, the cosine function is negative. 3. α = 225° – 180° = 45° 4. cos 945 ° = cos 225 ° = − cos 45 ° = −

1 2

g. 1. –930° = (–3 ⋅ 360°) + 150°. So the primary directed angle is 150° and it is in the second quadrant. 2. In the second quadrant, the tangent function is negative. 3. α = 180° – 150° = 30° 4. tan ( −930 °) = tan 1 50 ° = − tan 30° =−

1 3

h. 1. –675° = (–2 ⋅ 360°) + 45°. So the primary directed angle is 45° and it is in the first quadrant. 2. In the first quadrant, the cotangent function is positive. 3. α = 45° – 0° = 45° 4. cot (–675°) = cot 45° = 1 130

Algebra 10

EXAMPLE

46

Find the each trigonometric value by using a reference angle. a. cot

Solution

a. 1.

7π 6

b. tan

31π 4

⎛ 25π ⎞ c. sin ⎜ − ⎟ ⎝ 3 ⎠

⎛ 47 π ⎞ d. cos ⎜ − ⎟ ⎝ 4 ⎠

7 7π π 7π < 2 so = π + so the < 2π . So this is a primary directed angle. Moreover, 6 6 6 6 angle is in the third quadrant.

2. In the third quadrant, the cotangent function is positive. 3. α = 7π − π = π 6 6 4.

cot

7π π = cot = 3 6 6

31 > 2, we need to write the angle as a primary directed angle: 4 7π 31π 7π and it is in the fourth quadrant. = (3 ⋅ 2 π)+ . So the primary directed angle is 4 4 4 2. In the fourth quadrant, the tangent function is negative.

b. 1. Since

3. 4.

7π π = 4 4 π 31π 7π tan = tan = − tan = −1 4 4 4

α = 2π −

25 < −2, we need to write the angle as a primary directed angle. 3 25π 5π − = ( −5 ⋅ 2 π)+ . So the primary directed angle is 5π and it is in the fourth 3 3 3 quadrant.

c. 1. Since −

2. In the fourth quadrant, the sine function is negative. 5π π = 3 3

3.

α = 2π −

4.

π 5π 3 ⎛ 25π ⎞ sin ⎜ − = − sin = − ⎟ = sin 3 3 2 ⎝ 3 ⎠

d. 1.

47 π π 47 = −6 ⋅ (2 π)+ . < −2 so we need to find the primary directed angle: − 4 4 4 The angle is in the first quadrant. −

2. In the first quadrant, the cosine function is positive. π π 3. α = − 0 = 4 4 4.

π 1 ⎛ 47 π ⎞ cos ⎜ − ⎟ = cos = 4 2 ⎝ 4 ⎠

Trigonometry: Functions, Equations and Inequalities

131

Check Yourself 11 1. Find the reference angle for each angle. a. 890° b. 5000° c. –850° d. –2500° 50 π 32π 103π 11π e. f. g. − h. − 11 7 6 5 2. Find each trigonometric value by using a reference angle. a. sin 570°

b. tan 405°

19π 27 π f. cos 3 4 Answers 1. a. 10° b. 40° c. 50°

e. cot

2. a. −

1 2

b. 1

c.

3 2

c. cos (–2550°)

d. cot (–7950°)

g. sin ⎛⎜ − 45π ⎞⎟ ⎝ 4 ⎠

25π ⎞ h. tan ⎛⎜ − ⎟ ⎝ 6 ⎠

d. 20° d. –ñ3

3π 7 3 e. 3

e.

f.

π 6

f. −

2 2

π 5

g.

5π 11

h.

g.

2 2

h. −

3 3

3. Calculating Ratios from a Given Ratio In this section we will learn how to find all the trigonometric ratios of an angle from a single given ratio. In solving such problems we will use our knowledge of trigonometric ratios in right triangles, the sign of a trigonometric function and the fundamental trigonometric identities. Let us look at an example. EXAMPLE

47

Solution

For each trigonometric ratio in the given quadrant, find the five other trigonometric ratios for same angle. 1 2 a. sin θ = , θ ∈ (0 °, 90 °) b. cos θ = − , θ ∈ (90 °, 180 °) 3 5 3π c. tan θ = 7 , θ ∈ ⎛⎜ π, 3π ⎞⎟ d. cot θ = −6, θ ∈ ⎛⎜ , 2 π ⎞⎟ 4 ⎝ 2 ⎠ ⎝ 2 ⎠ We will use the abbreviations opp, adj and hyp to mean the opposite side, adjacent side and hypotenuse of a triangle with respect to the angle θ. a. The angle is in the first quadrant. In this quadrant, both axes are positive and so the sine and cosine values will be positive. By the Pythagorean Theorem, (+) hyp2 = adj2 + 22 2 2 2 5 = adj + 2 5 adj2 = 25 – 4 +2 q q

adj = ±ò21.

(+)

q

+x

We choose the positive value for the first quadrant: adj = ò21. As a result, cos θ = 132

21 2 21 5 5 , tan θ = , cot θ = , sec θ = , csc θ = . 5 2 2 21 21 Algebra 10

b. The angle is in the second quadrant. In this quadrant

(+)

the x-axis is negative and the y-axis is positive. The q

cosine function is related to the x-axis, and so cos θ = −

3

+y

qq –1

(+)

1 adj −1 can be taken as cos θ = = . 3 hyp +3

By the Pythagorean Theorem, opp = ñ8 = 2ñ2. As a result, sin θ =

3 3 2 2 2 2 −1 = −3, csc θ = . , tan θ = = −2 2, cot θ = , sec θ = −1 2 2 3 −1 2 2

opp 7 −7 using the signs of = = adj 4 −4 the axes in the third quadrant. By the Pythagorean

c. Similarly, tan θ =

Theorem, hyp = ò65. As a result, sin θ = sec θ =

−7 65

, cos θ =

−4 65

, cot θ =

–4 (–)

q

(+)

−4 4 = , −7 7

q

–7

r

(–)

65 65 , csc θ = . −4 −7

adj 6 = −6 = using the signs of the opp −1 axes in the fourth quadrant.

d. Finally, cot θ =

By the Pythagorean Theorem, hyp = ò37. As a result, sin θ = tan θ =

−1 6 , cos θ = , 37 37

(+) q

q

+6 –1

r (–)

37 csc θ = 37 = − 37. −1 , , sec θ = −1 6 6

Trigonometry: Functions, Equations and Inequalities

133

Check Yourself 12 1. Find the sine, cosine and tangent ratios of each angle without using a trigonometric table or calculator. 2π 3 2. a. tan θ = 5, θ ∈ (180°, 270°) are given.

a. θ = 315°

c. θ = – 900°

b. θ =

d. θ =

63π 2

Find sin θ and cos θ. b. sec θ = −10, θ ∈ ⎛⎜ π , π ⎞⎟ are given. ⎝2 ⎠ Find sin θ and tan θ. Answers 1. a. sin 315 ° = − b. sin

1

2π 3 = , 3 2

c. sin 900° = 0, d. sin

63π = −1, 2

2. a. sin θ = − b. sin θ =

5 26

3 11 10

2

,

cos 315 ° =

cos

1 2

,

2π 1 =− , 3 2

cos 900° = –1, cos

63π = 0, 2

cos θ = −

tan 315° = –1 tan

2π =− 3 3

tan 900° = 0 tan

63π is undefined 2

1 26

tan θ = –3ò11

C. PERIODS OF TRIGONOMETRIC FUNCTIONS 1. Periodic Functions The values of some mathematical functions repeat with the same pattern to infinity. A function with this property is called a periodic function. Definition

periodic function, period, fundamental period A function f: A → B is a periodic function if it satisfies the rule f(x + T) = f(x) for all values of x in A. The number T is called the period of the function. The smallest possible positive value of T is called the fundamental period of the function.

134

Algebra 10

EXAMPLE

48

Solution

f : → is a function such that f(x) = {the remainder when x is divided by 4}. Draw the graph of f and find its period. Let us find some values of f(x) to draw the graph. x 0 1 2 3 4 5 6 7 8 9 10

f(x) 0 1 2 3 0 1 2 3 0 1 2

y = f(x) 3 2 1 1 2 3 4 5 6 7 8 9 10

x

As we can see in the table, the values 0, 1, 2 and 3 repeat. So we can write f(x) = f(x + 4) = f(x + 8) = … . We can say that T = 4, 8, 12, ... are all periods of f(x). The smallest positive period of f(x) is 4, so the fundamental period of f(x) is 4.

EXAMPLE

49

The figure below shows the graph of the function y = f(x). a. Find the fundamental period of f(x). b. Find f(133), f(85) and f(202). y = f(x) 2 –5 –4

5 –3 –2

1 2 3 4

10 6 7 8 9

12 11

15 13 14

x

–2

Trigonometry: Functions, Equations and Inequalities

135

Solution

a. From the graph we can see f(–5)= f(0) = f(5) = ... = –2 f(–4)= f(1) = f(6) = ... = 0 f(–3)= f(2) = f(7) = ... = 2 f(–2)= f(3) = f(8) = ... = 2 … So the periods are 5, 10, 15, ... and the fundamental period is T = 5. b. If T = 5 then we can write f(x) = f(x + kT) = f(x + 5k), k ∈ . So f(133) = f(3 + 26 ⋅ 5) = f(3) = 2 f(85) = f(0 + 17 ⋅ 5) = f(0) = –2 f(202) = f(2 + 40 ⋅ 5) = f(2) = 2.

2. Periods of Trigonometric Functions We know that the tangent and cotangent values of angles in the first and third quadrants and in the second and fourth quadrants are the same, so we only add kπ for the tangent and cotangent functions.

EXAMPLE

50

Solution

Recall the definition of coterminal angles on the unit circle: for any angle x we can write x = x + 2kπ, k ∈ . If we apply this fact to trigonometric functions, we get sin x = sin(x + 2kπ) cos x = cos(x + 2kπ) tan x = tan(x + kπ) cot x = cot(x + kπ). This means that trigonometric functions are also periodic functions. Moreover, 2kπ is a period for sin x and cos x and kπ is a period for tan x and cot x, where k is an integer.

Find the fundamental period of f(x) = 2cos 3x. If f is a periodic function then by definition, f(x) = f(x + T) . So 2 cos 3x = 2 cos [3(x + T)]. Simplifying gives us cos 3x = cos [3(x + T)].

If cosx = cosy then x = ±y + 2kπ. When we write –y + 2kπ, we can simplify 3x from both sides. For this reason, we will take only the positive value of y.

136

Let us use the coterminal angles of 3x, i.e 3x + 2kπ: cos(3x + 2kπ) = cos [3(x + T)]. Removing the cosine function from both sides gives 3x + 2kπ = 3(x + T) = 3x + 3T, so 2kπ is the period of f(x). 3 2π The smallest possible positive value of k is 1, and so the fundamental period of f(x) is . 3

2kπ = 3T and therefore T =

Algebra 10

Rule

Let n be an integer, let a and b be real numbers and let T stand for the fundamental period of a trigonometric function. 1. For y = sin n( ax + b), y = cos n( ax+ b), y = sec n( ax+ b)and y = csc n( ax+ b). i. if n is even then T = ii. if n is odd then T =

π . | a|

2π . | a|

2. For y = tan n( ax + b) and y = cot n( ax+ b), T =

π . | a|

Note If f(x) is a periodic function then af(x) and f(x) + b are also periodic functions and the periods of all these functions are the same. Can you see why?

EXAMPLE

51

Find the fundamental period of each function. a. y = sin7 8x.

b. y = 2cos5 (4x + 2).

c. f(x) = 3 sin2 (3x + 5) – 2.

d. f(x) = 7 sec3 (5 – 2x) + 4.

e. f(x) = csc4 (4x + 1) – 1. Solution

We will use the rule we have just seen. a. n = 7 is odd, so T =

2π 2 π π = = . | a| 8 4

b. n = 5 is odd, so T =

2π 2 π π = = . | a| 4 2

c. n = 2 is even, so T =

π π = . | a| 3

d. n = 3 is odd and a = – 2, so T = e. n = 4 is even, so T =

EXAMPLE

52

2π 2π 2π = = =π. | a | | –2| 2

π π = . | a| 4

Find the fundamental period of each function. a. y = 5 – cot3 (2x – 1).

Trigonometry: Functions, Equations and Inequalities

b. f(x) = tan4 5x.

c. y = 2 cot3 (2 – 6x) + 1. 137

Solution

a. We know that the period of y = 5 – cot3(2x – 1) is the same as the period of π π y = cot3(2x – 1). So T = = . | a| 2 π π b. T = = | a| 5 c. a = –6, so T =

EXAMPLE

53

Solution

π π π = = . | a | | –6| 6

π Find the fundamental period of y = sin( x). 3

n = 1 is odd and a = T=

π , so 3

2π 2 π 3 = = 2π ⋅ = 6 . | a| π π 3

Rule

least LCM means common multiple: the smallest quantity that is divisible by two or more given quantities without a remainder. GCD is the greatest common divisor: the greatest quantity that can divide two or more given quantities.

To find the period of the sum or difference of two or more periodic functions, first we find the fundamental period of each separate function, and then we take the least common multiple of these periods. This is the fundamental period of given function. For example, if f(x) = g(x) + h(x) and the fundamental periods of g(x) and h(x) are T1 and T2 respectively then the fundamental period of f(x) is T = LCM(T1, T2).

For example: LCM (4,6) = 12, GCD (4,6) = 2.

Rule

For any two quantities

a c and , b d

a c LCM( a, c) LCM( , ) = b d GCD( b, d) a c p LCM( a, c, ..., p, ...) By extension, LCM( , , ..., , ...) = . b d q GCD( b, d, ..., q, ...)

EXAMPLE

138

42

Find the fundamental period of y = sin3x + cos33x. Algebra 10

Solution

For g(x) = sin3x we have T1 = 2π. For h(x) = cos33x we have T2 =

2π . 3

So the fundamental period is T = LCM(2 π, EXAMPLE

54

Solution

Find the fundamental period of y = 3 cos5 3x – 2 cot3 5x. 2π . 3 π For h( x) = 2 cot 3 5 x we have T2 = . 5

For g(x) = 3 cos5 3x we have T1 =

So the period is T = LCM(

EXAMPLE

55

Solution

2π 2π 2π LCM(2 π, 2 π) 2 π ) = LCM( , )= = = 2 π. 3 1 3 GCD(1, 3) 1

2π π LCM(2 π, π) 2 π , )= = = 2 π. 3 5 GCD(3, 5) 1

Find the fundamental period of y = cos2 3x – cot2 2x + sin3(5x + 1). π For g( x) = cos 2 3 x we have T1 = , 3 π for h( x) = cot 2 2 x we have T2 = , 2 and for r( x) = sin 3(5 x+1) we have T3 =

2π . 5

π π 2π LCM( π, π, 2 π) 2 π So T = LCM( , , )= = = 2 π. 3 2 5 GCD(3, 2, 5) 1

Note If f(x) is a product or quotient of two periodic functions then LCM(T1, T2) gives us a period of f(x) but this may not be the fundamental period. To find the fundamental period we must change the product or quotient to a sum or difference of functions. EXAMPLE

56

Solution

Find the fundamental period of y = cos 5x ⋅ cos 3x. First rewrite the product as a sum or difference: y = cos 5x ⋅ cos 3x = No ow for g( x) =

1 1 1 ⋅ [cos(5x + 3x)+ cos(5x 3x)] = cos 8 x + cos 2 x. 2 2 2

1 2π π 1 2π cos 8 x we get T1 = = , and for h( x) = cos 2 x we get T2 = = π, so 2 8 4 2 2

π π π LCM( π, π ) π = = π. T = LCM( , π) = LCM( , ) = 4 4 1 GCD( 4, 1) 1 Trigonometry: Functions, Equations and Inequalities

139

Check Yourself 13 1. Find the fundamental period of each function. a. y = sin3 4x

b. y = 2 sin2(4 – 3x)

c. y = 3 cos3(5x – 2)

d. y = 2 sec2(2 – 4x) + 1

e. y = csc7(2x + 3) – 4

f. y = 2 cos 5x

2. Find the fundamental period of each function. a. y = cot6(3x + 2)

b. y = tan 3x

c. y = 2 cot 3(3 –

π 3. Find the fundamental period of y = sin( x) . 5

2πx ) 5

d. y = tan

x 4

4. Find the fundamental period of each function. a. y = sin3 2x + cos3 5x

b. y = cos2 4x – 2 cot5 3x

c. y = cos3 8x – cot 2x + sin3(6x + 1) d. y = cos x ⋅ cos 3x e. y = 6 sin 2x ⋅ cos 2x

f. y = sin x ⋅ sin 3x

Answers 1. a.

π 2

b.

π 3

c.

2π 5

d.

π 4

4. a. 2π b. π c. π d. π e.

e. π f. π 2

2π 5

2. a.

π 3

b.

π 3

c.

5 2

d. 4π

3. 10

f. π

D. GRAPHS OF TRIGONOMETRIC FUNCTIONS We have already seen that trigonometric functions are periodic. We can use this fact to sketch or draw the graphs of trigonometric functions. To draw the graph of a trigonometric function, follow the steps: 1. Find the period of the function. 2. Choose a suitable interval for the graph. Generally [0, 2π] is a good interval for sin x and π π cos x, (0, π) is good for cot x and (– , ) is good for tan x. 2 2 3. Investigate the behavior of the function in the given interval. For this we will check the values of the function for common angles such as 0°, 30°, 45°, 60°, 90°, ... . This will show us when the function is increasing and when it is decreasing. We write to show that a function is increasing on an interval. We write to show that it is decreasing. For sin x and cos x functions we can divide the interval [0, 2π] into four equal parts and take the endpoints of the intervals as the angles to inspect. 4. Draw the graph of the function in the given interval and copy it any number of times to get the general graph of the function. Now we are ready to look at the graph of the basic trigonometric functions. 140

Algebra 10

1. Graph of the Sine Function → [–1, 1]. Follow the steps described above:

For f(x) = sin x we have f: 1. The period of sin x is 2π.

2. We will draw the graph on the interval [0, 2π]. 3. We can use the quadrantal angles 0,

x sin x

π 3π , π, and 2 π and make a table: 2 2

0

p 2

p

3p 2

2p

0

1

0

–1

0

4. Now we can draw the graph with the information. y

1 –p

p 2

3p 2

p

3p 2p

x

–1 2p

Remember that the period of y = sin x is 2π, so after drawing the graph in [0, 2π] we can copy the curve any number of times to get the general graph of y = sin x.

2. Graph of the Cosine Function For f(x) = cos x we have f:

→ [–1, 1].

1. The period of cos x is 2π. 2. We will draw the graph on the interval [0, 2π]. 3. We can use the quadrantal angles π 3π 0, , π, and 2 π and make a table: 2 2

x cos x

0

p 2

p

3p 2

2p

1

0

–1

0

1

4. Now we can draw the graph: y 1

–

Trigonometry: Functions, Equations and Inequalities

3p 2

–p –

p 2

–1

y = cos x p 2

p

3p 2

2p

x

2p

141

The period of y = cos x is 2π, so after drawing the graph in [0, 2π] we can copy the curve any number of times to get the general graph of y = cos x.

3. Graph of the Tangent Function f(x) = tan x means f :

π – { + kπ} → 2

.

1. The period of tan x is π. π π 2. We will draw the graph on the interval (– , ) . 2 2 π π π π 3. Use the special angles − , − , 0, and to make a table: 2 4 4 2 x

–

p 2

–

tan x –¥

p 4

0

p 4

p 2

–1

0

1

¥

4. Draw the graph:

y

– 3p 2 –p

–p 2

p 2 p

3p 2

x

π π The period of y = tan x is π, so after drawing the graph on the interval (– , ) we can 2 2 copy the curve any number of times to get the general graph of y = tan x.

4. Graph of the Cotangent Function f(x) = cot x means f :

– {kπ} →

.

1. The period of cot x is π. 2. We will draw the graph on the interval (0, π). π π 3π π 3. Use the special angles 0, , , to make a table: and 4 2 4 2 x cot x

142

0

p 4

p 2

3p 4

p

¥

1

0

–1

–¥

Algebra 10

4. Draw the graph:

y

–p

–p 2

3p 2

p 2 p

x

2p

The period of y = cot x is π, so after drawing the graph in (0, π) we can copy the curve any number of times to get the general graph of y = cot x.

5. Graph of the Secant Function 1 . So f(x) = sec x is undefined when cos x = 0, which means cos x π the function is f : – { + kπ} → – (–1, 1). 2 1. The period of sec x is 2π.

We know that sec x =

2. We will draw the graph on the interval (0, 2π). 3. We can use the quadrantal angles and the cosine function to make the table: x sec x

0

p 2

p

3p 2

2p

0

+¥ –¥

–1

–¥ +¥

1

4. Draw the graph:

y y = sec x

1 -3p/2 -2p

p/2

-p/2 -p

-1

3p/2 p

2p

x y = cos x

Remember that the period of y = sec x is 2π, so we can copy the curve any number of times to get the general graph of y = sec x. Trigonometry: Functions, Equations and Inequalities

143

6. Graph of the Cosecant Function We know that f(x) = csc x and csc x = sin x = 0, i.e. the function is f :

1 , so we can say that csc x is undefined when sin x

– {kπ} →

– (–1, 1).

1. The period of csc x is 2π. 2. We will draw the graph on the interval (0, 2π). 3. We can use the quadrantal angles and the sine function to make table:

x

0

csc x +¥

p 2

p

3p 2

2p

1

+¥ –¥

–1

–¥ +¥

4. Draw the graph: y y = csc x

1 -p

p

-p/2

-3p/2

O -1

p/2

3p/2 2p

x

5p/2 y = sin x

Remember that the period of y = csc x is 2π, so after drawing the graph in [0, 2π] we can copy the curve any number of times to get the general graph of y = csc x.

7. Graph Translations Recall the basic effects that different changes in a function can have on a graph: 1. The graph of y = f(x – r) is the graph of f(x) shifted r units right along the x-axis. 2. The graph of y = f(x) + k is the graph of f(x) shifted k units up along the y-axis. 3. The graph of y = –f(x) is the reflection of the graph of f(x) with respect to the x-axis. 144

Algebra 10

EXAMPLE

57

Solution

π Draw the graph of y = sin( x – ) . 3 π π , so we will shift the graph of f(x)= sin x to the right by units along the x-axis. So 3 3 π the graph of y = sin( x – ) is as follows: 3 r=

y y = sin x

1 p 3

–p

y = sin (x –

2p

p

p ) 3 x

–1

EXAMPLE

58

Solution

Draw the graph of y = 3cos(2x + 1). Let us draw the graph of y = 3cos 2x first, then we will shift the graph one unit to the left. 2π 1. The period of y = cos 2x is T = =π. 2 2. We will draw the graph in [0, π]. π π 3π and π to 3. Use the special angles 0, , , 4 2 4 make the table:

x

0

p 4

p 2

3p 4

p

cos 2x

1

0

–1

0

1

3cos 2x

3

0

–3

0

3

4. Now we can draw y = cos 2x and then y = 3cos (2x + 1): y y = 3cos (2x + 1)

3

y = 3cos 2x

p 4

–3

Trigonometry: Functions, Equations and Inequalities

p 2

3p 4

p

3p 2

2p

x

p

145

EXAMPLE

59

Solution

Draw the graph of y = 2 sin 3x + 4. Begin with the graph of y = 2 sin 3x: 1. The period of y = 2 sin 3x is T =

2π . 3

2π ]. 3 π π π 2π 3. Use the angles 0, , , , and to draw the table: 6 3 2 3

2. We will draw the graph in [0,

x

0

p 6

p 3

p 2

2p 3

sin 3x

0

1

0

–1

0

2sin 3x

0

2

0

–2

0

2sin 3x + 4

4

6

4

2

4

4. Finally, draw the graph:

y 6 4 2 p p p 2p 6 3 2 3

x

2p 3

EXAMPLE

60

Solution

x – 3 in the interval [–6π, 6π]. 3 x 2π 1. The period of y = cos is T = = 6 π. 1 3 3

Draw the graph of y = 2 cos

3π 9π 2. Let us divide the interval [0, 6π] into four equal parts, then the angles will be 0, , 3π, , 2 2 6π.

3. Make the table: x

0

p cos 1 3 p 2cos – 3 –1 3

146

3p 2

3p

9p 2

6p

0

–1

0

1

–3

–5

–3

–1

Algebra 10

4. Finally, draw the graph: y –6p

–

9p 2

–3p

–

3p 2

3p 2

3p

9p 2

6p x

–1 –3 –5

EXAMPLE

61

Solution

x in [–3 π, 3 π]. 3

Draw the graph of y = tan 1. The period of y = tan

6p

π x is T = = 3 π. 1 3 3

2. We need to draw the graph in [–3π, 3π]. 3. Let us use the angles – x tan

3p 2

0

3p 2

–¥

0

¥

–

x 3

3π 3π and draw the table: , 0, 2 2

4. Now we can draw the graph and repeat the curve on the interval [–3π, 3π]: y

3p 2

– 3p 2 –3p

3p

x

3p

Trigonometry: Functions, Equations and Inequalities

147

Check Yourself 14 1. Draw the graph of each function in the interval [0, π]. a. y = 3 sin 4x – 1

b. y = 5 – cos 2x

c. y = tan 2x

b.

c.

Answers 1. a. 2

–1

p 8

p 3p p 4 8 2

3p 4 5p 8

p

6 5 4

7p 8

–4

p 4 p 4

p 2

3p 4

p 2

3p 4

p

p

E. INVERSE TRIGONOMETRIC FUNCTIONS Remember! If f: D → R is a one-to-one and onto function then : R → D is its inverse.

f

–1

one function if for every x1 ≠ x2, f(x1) ≠ f(x2). Also, Recall that a function f: S → T is a one-tto-o f is an onto function if every element in T is an image of an element in S. If a function is both one-to-one and onto then it has an inverse. It is easy to show that trigonometric functions are not one-to-one, since they repeat themselves. Therefore, if we want to define the inverse of a trigonometric function, we must choose a suitable smaller domain and range in which the function is both one-to-one and onto. We use the prefix ‘arc’ to name the inverse of a trigonometric function. For example, arcsin x is the inverse of sin x, arccos x is the inverse of cos x, arctan x is the inverse of tan x, arccot x is the inverse of cot x, and so on.

Note

Be careful! sin −1 x ≠

1 sin x

cos −1 x ≠

1 cos x

Some books use sin–1 x, cos–1 x, tan–1 x, cot–1 x or Asin x, Acos x, Atan x, Acot x to show the inverses of trigonometric functions. Some book also use Arcsin x to mean the inverse function of sin x, and arcsin x to mean the inverse relation of sin x. We know that a relation does not need to be one-to-one to have inverse, but a function must be one-to-one and onto to have its inverse. In this book we will use arcsin x, arccos x, arctan x and arccot x to mean the inverses of the trigonometric functions.

148

Algebra 10

1. The Arcsine Function ⎡ π π⎤ If f : ⎢ − , ⎥ → [–1, 1] then f(x) = sin x is both one-to-one and onto, so we can define its ⎣ 2 2⎦ inverse on this interval. The inverse of f(x) = sin x on this interval is f –1(x) = arcsin x ⎡ π π⎤ with f −1 : [ −1, 1] → ⎢ − , ⎥. ⎣ 2 2⎦

π π x = arcsin y ⇔ y = sin x, x ∈[ − , ] 2 2

Graph of y = arcsin x ⎡ π π⎤ For arcsin : [ −1, 1] → ⎢ − , ⎥ we can find the following values: ⎣ 2 2⎦ Remember! The graph of an inverse function f –1(x) is the reflection of the graph of the function f(x) with respect to the line y = x.

x

–

–1

arcsin x – p 2

ñ2 2

–

0

p 4

0

ñ2 2

1

p 4

p 2

The values give us the graph y = arcsin x: y p y = arcsin x 2 1 y = sin x –

p 2

–1 0

y=x

EXAMPLE

62

Solution

1

p 2

x

–1 p – 2

1 Find the value of arcsin . 2

By the definition of inverse trigonometric functions we have 1 1 x = arcsin ⇔ = sin x . 2 2 1 We know that sin 30° = sin 150° = sin 390° = … = , but arcsin is only defined on the 2 π ⎡ π π⎤ interval ⎢ − , ⎥ , so we will take x = 30° = . 6 ⎣ 2 2⎦

Trigonometry: Functions, Equations and Inequalities

149

EXAMPLE

63

Solution

EXAMPLE

64

Find arcsin(–

3 ). 2

⎡ π π⎤ The arcsine function is defined on the interval ⎢ − , ⎥ , ⎣ 2 2⎦

so x = arcsin(–

3 3 π )⇔– = sin x, i.e. x= –60 °= – . 2 2 3

Find sin(arcsin

2 ). 2

Let arcsin

2 2 = x . Then we can write sin(arcsin ) = sin x . 2 2

Remember! By the property of inverse functions,

x = arcsin

2 2 ⇔ = sin x, so x = 45 °. 2 2

f(f –1(x)) = x and

So sin(arcsin

Solution 1

–1

f (f (x)) = x.

Solution 2

EXAMPLE

65

Solution

sin(arcsin

2 2 ) = sin x = sin 45 °= . 2 2

2 2 by the property of inverse functions (since f(f –1(x))=x). )= 2 2

Find cos (arcsin 1). Let arcsin 1 = x, then we must find cos x. If arcsin 1 = x then sin x = 1 and so x = 90°. So cos (arcsin 1) = cos x = cos 90° = 0.

EXAMPLE

66

Solution

150

Find arcsin (tan 45°). π tan 45° = 1, so we need to find arcsin1= 90 ° = . 2 Algebra 10

2. The Arccosine Function If f: [0, π] → [–1, 1] then f(x) = cos x is both one-to-one and onto, so we can define its inverse. The inverse of f(x) = cos x on this interval is f –1(x) = arccos x with f –1: [–1, 1] → [0, π]. x = arccos y ⇔ y = cos x, x ∈ [0, π]

Graph of y = arccos x For arccos : [–1, 1] → [0, π] we can find the following values:

x arccos x

–

–1 p

ñ2 2

0

ñ2 2

3p 4

p 2

p 4

1 0

The values give us the graph of the function y = arccos x: y y = arccos x

Notice that the graph is a reflection of

p

y=x

y = cos x in the line y = x.

p 2 1

0

–1 –1

EXAMPLE

67

Solution

1 p 2

p x y = cos x

1 Find arccos . 2

By the definition of inverse trigonometric functions we have 1 1 1 π 1 x = arccos ⇔ = cos x and x ∈ [0, π] ⋅ = cos x means x = 60° = = arccos . 2 2 2 3 2

Trigonometry: Functions, Equations and Inequalities

151

EXAMPLE

68

Solution

Find arccos

3 2 + arccos . 2 2

Let us evaluate each term separately. a = arccos

3 3 π means cos a = , so a= 30 °= . 2 2 6

b = arccos

2 2 π means cos b = , so b= 45 °= . 2 2 4

π π 5π Combining these results gives us a+ b= + = . 6 4 12 EXAMPLE

69

Solution

Find arccos (cos (–30°)). Since cos x and arccos x are inverse functions, the answer seems to be –30°. But we know arccos: [–1, 1] → [0, π], so the result cannot be –30°. Instead, we must find the answer in [0, π]. If cos (–30°) = a then we must find arccos a. Since cos (–x) = cosx, we have a = cos(–30 °) = cos 30 ° =

3 . 2

So arccos(cos(–30°)) = arccos a= arccos

EXAMPLE

70

Solution

Find sin(arcsin

Let a = arcsin

2 3 + arccos ). 2 2 2 3 and b = arccos . 2 2

a = arcsin

π 2 2 means sin a = and so a= 45 ° = . 2 2 4

b = arccos

π 3 3 means cos b = and so b= 30 ° = . 2 2 6

So sin(arcsin

2 3 + arccos ) = sin(45° + 30°) = (sin45° ⋅ cos30°) + (cos45° ⋅ sin30°) 2 2 =(

152

π 3 = 30 °= . This is the answer. 2 6

2 3 2 1 6+ 2 ⋅ ⋅ )= )+( . 2 2 2 2 4 Algebra 10

EXAMPLE

71

Solution

3 Find cos(arcsin ). 5 3 3 Let a = arcsin , then sin a = and we 5 5 3 need to find cos a. However, sin a = is not 5 a ratio we know, so we can draw a right

A

5

3

triangle to find the other trigonometric ratios for the angle a. In the figure, the Pythagorean 2

2

2

Theorem gives us BC + AB = AC , i.e.

a 4

C

B

BC2 = 52 – 32 = 25 – 9 = 16, so BC = 4. BC 4 3 So cos a = = = cos(arcsin ). AC 5 5

EXAMPLE

72

Solution

π 1 Find cos( – arcsin ). 2 2 1 Let arcsin = a, then we need to find 2 cos(

Reduction formula: π cos( − x) = sin x 2

EXAMPLE

73

Solution

1 1 π – a) = sin a = sin(arcsin ) = . 2 2 2

2 Find sin(2 ⋅ arccos ). 3

2 2 = a, then cos a = . 3 3 Then we have 2 sin(2 ⋅ arccos ) = sin 2a = 2 sin a ⋅ cos a. 3 2 Given that cos a = we can draw a right 3 triangle to find sin a.

A

Let arccos

The Pythagorean Theorem gives us AC = 5, so sin a =

3

ñ5

a B

2

C

5 . So 3

2 sin(2 ⋅ arccos ) = sin 2 a = 2 ⋅sin ⋅ a ⋅cos a 3 5 2 4 5 ⋅ = =2⋅ . 3 3 9 Trigonometry: Functions, Equations and Inequalities

153

EXAMPLE

74

Solution

Find the domain of y = arccos(2x + 1). arccos: [–1, 1] → [0, π], so –1 ≤ (2x + 1) ≤ 1, and subtracting 1 from all parts of the inequality gives us –2 ≤ 2x ≤ 0. Finally, dividing all parts by 2 gives us –1 ≤ x ≤ 0. So x ∈ [–1, 0], i.e. the domain is [–1, 0].

3. The Arctangent Function ⎛ π π⎞ If f : ⎜ − , ⎟ → ⎝ 2 2⎠

then f ( x) = tan x is both one-to-one and onto, so we can define its

inverse. The inverse of f(x) = tan x is f –1(x) = arctan x with f −1 :

⎛ π π⎞ → ⎜− , ⎟ . ⎝ 2 2⎠

⎛ π π⎞ x = arctan y ⇔ y = tan x, x ∈ ⎜ − , ⎟ ⎝ 2 2⎠

Graph of y = arctan x ⎛ π π⎞ For arctan : ⎜ − , ⎟ → ⎝ 2 2⎠ x –¥

–1 p – 4

arctan x – p 2

we can find the following values: 0

1

¥

0

p 4

p 2

The values give us the graph of the function y = arctan x: y y = tan x y=x y = arctan x

p – 2

0 p 2 –

EXAMPLE

75

Solution 154

x

p 2

Find arctan ñ3. If x = arctan ñ3 then tan x = ñ3 and so x = 60° =

π = arctan 3. 3 Algebra 10

EXAMPLE

76

Solution

Find arctan1+ 2 ⋅ arctan

3 – arctan(– 3). 3

If arctan 1 = a then tan a = 1 and so a = 45° = If arctan

π . 4

3 3 π = b then tan b = and so b= 30 ° = . 3 3 6

π If arctan(– 3) = c then tan c= – 3 and so c= –60 ° = − . 3

So arctan 1 + 2 ⋅ arctan

EXAMPLE

77

Solution

π π π 11π 3 . – arctan (–ñ3 ) = + 2 ⋅ − ( − ) = 4 6 3 12 3

1 Find sin(arctan ). 2 1 1 1 If arctan = x then tan x = and we need to find sin x. However, tan x = is not a 2 2 2 ratio we know, so we can draw a right triangle to find other trigonometric ratios of x.

In the figure, the Pythagorean Theorem gives us BC2 + AC2 = AB2, i.e.

A ñ5

AB2 = 22 + 12 = 4 + 1 = 5 and so AB = ñ5. So sin x =

EXAMPLE

78

Solution

1 5 AC = = . 5 AB 5

Find arctan(tan

x B

2

1 C

5π ). 4

By the properties of inverse functions the answer seems to be

5π , but remember that 4

5π ⎛ π π⎞ cannot be the answer. arctan a ∈ ⎜ − , ⎟ and so 4 ⎝ 2 2⎠ 5π π However, tan =1 and so arctan 1= 45 ° = is the correct answer. 4 4

4. The Arccotangent Function If f: (0, π) → then f(x) = cot x is both one-to-one and onto, so we can define its inverse. The inverse of f(x) = cot x is f –1(x) = arccot x with f –1: → (0, π). x = arccot y ⇔ y = cot x, x ∈ (0, π) Trigonometry: Functions, Equations and Inequalities

155

Graph of y = arccot x For arccot: (0, π) → x –¥ p

arccot x

we have the following values:

–1

0

1

¥

3p 4

p 2

p 4

0

The values give us the graph of arccot x: y

y=x

p

y = arccot x p 2

0

p 2

p

x

y = cot x

EXAMPLE

79

Solution

EXAMPLE

80

Solution

Find arccot x = arccot

3 . 3

3 3 π means cot x = , so x= 60 °= . 3 3 3

Find tan(arccot ñ3) + cos (arccot 1) – sin (arccot 0). If a = arccot ñ3 then cot a = ñ3 and so a = 30°. If b = arccot 1 then cot b = 1 and so b = 45°. If c = arccot 0 then cot c = 0 and so c = 90°. So tan (arccot ñ3) + cos (arccot 1) – sin (arccot 0) = tan 30° + cos 45° – sin 90° =

156

3 2 2 3+3 2 −6 + −1 = . 3 2 6 Algebra 10

Check Yourself 15 1. Find each value in radians. 1 b. arcsin(– ) 2

a. arcsin 1 2 ) 2

d. arccos(–

1 c. arccos(– ) 2 f. arctan(– 3 ) 3

e. arctan 0

g. arccot ñ3

h. arccot (–1)

2. Evaluate each expression. 2 ) 2

a. cos(arcsin

3 ) 2

b. sin (arccos 1)

c. tan(arcsin

e. arccot (tan 60°)

f. arccos (cot 45°)

2 a. sin(arcsin ) 5

3 b. cos(arcsin ) 4

2 c. tan(arcsin ) 3

d. cot(arctan 3)

e. sin (arctan 2)

f. cos (arcsin(cos 60°))

d. sin(arctan(–

3 )) 3

3. Evaluate each expression.

4. Evaluate each expression. 3π 3 + arctan ) 2 4

3 a. sin( π − arcsin ) 5

b. tan(

2 d. cos(2 ⋅ arcsin ) 7

1 e. tan(arccot ) 4

π 3 c. cot( – arcsin ) 2 2

f.

1 tan(2 ⋅ arcsin ) 2

5. State the domain of each function. a. f(x) = arcsin(3x – 1)

b. f(x) = arccos(4x + 2)

Answers 1. a. 2. a.

π 2 2 2

b. −

π 6

c. 2π 3

b. 0 c. ñ3 d. – 7 4

3. a.

2 5

b.

4. a.

3 5

b. –

⎡ 2⎤ 5. a. ⎢0, ⎥ ⎣ 3⎦

4 3

c.

2 5 5

c. ñ3 d.

d. 3π 4

e. 0

π 6

f. 0

1 2

d. 41 49

e. 1 3

e.

2 5 5

f.

f. −

π 6

g. π 6

h.

3π 4

3 2

e. 4 f. ñ3

1⎤ ⎡ 3 b. ⎢ – , – ⎥ 4 4 ⎣ ⎦

Trigonometry: Functions, Equations and Inequalities

157

MATHEMATICS AND MUSIC Many people today think that mathematics is a science, made up only of theorems, numbers and formulas. However, we can also think of mathematics as an art, just like painting or music. If you study the history of music or art, you will find that in ancient times mathematics and geometry were used to produce beautiful creative works. Also, mathematicians look for the simplest or most elegant proofs of a theorem, just as musicians try to produce the most beautiful or pleasing music. In ancient Greece the study of mathematics had four main parts: arithmetic, geometry, astronomy and music. The famous philosopher Confucius (551-479 BC) investigated the effects of musical notes and rhythms on humans, and Plato (428-348 BC) accepted music as an important part of education. Even in the Middle Ages (around 400 to 1500 AD), mathematics, astronomy and music were all taught together. Many famous mathematicians in history have also studied the relationships between mathematics and music. Music, Rhythm and Ratio One very basic relationship between math and music is rhythm. A musical rhythm can be defined as a mathematical ratio such as 4:4 or 3:4. With only a little practice, you can hear this ‘rhythm ratio’ in different pieces of music. Ratios also define musical notes. Pythagoras (580-520 BC) and his followers studied the relationship between the length of a piece of wire and the sound (or musical note) it makes, and saw that the relationships were ratios. For example, imagine that a piece of wire makes the note C (‘do’) when you pluck it. Pythagoras found that if you divide the string into the ratio 2:3 and pluck the longer part, the note is A (‘la’). If you divide the longer part into two-thirds again, the next note is F (‘fa’) in the next octave. Each time you divide the string, it produces a note which is five notes ahead in the musical scale. Other important mathematical ratios in the Pythagorean musical scale are 6:5, 3:2 and 16:9. You can find another link between music and math in the shapes of instruments. Many instruments have curves that are designed to give the best sound, and some of these curves correspond to the graph of y= 2x.

Sound Waves Each sound that you hear, such as a musical note, can be represented as a sound wave with a particular frequency and amplitude. Higher sounds have higher frequencies, and louder sounds have bigger amplitude. You may remember that a famous ratio in mathematics is the golden ratio, which is equal to 1+ 5 1.618 . 2 Musicians have found that if the frequencies of two notes are in this ratio then they produce a pleasing harmony when they are played together. Sound waves are connected to trigonometry by the sine function: every sound can be represented as a wave which has the equation y = a ⋅ sin(bθ – c). In this equation, a is the amplitude, b is the product of the frequency and 2π, and c is a real number called the phase difference. As an example, consider a sound with frequency 55 Hz and amplitude 3. The graph of this sound wave is given by y = a ⋅ sin(bθ – c) with a = 3, b = 2π ⋅ 55 = 110π and c = 0, i.e. y = 3sin (110πθ).

Math and Music We can see that mathematics provides one way for us to understand music and harmony. If a musician uses mathematical ratios in his music, the music will sound pleasing to more people. However, a musician does not need to know math to write music, and the people who listen to the music do not usually need any special education to enjoy it. This is one difference between music and most of mathematics. Can you think of any others?

Activity Try looking for pictures of sound waves in an encyclopedia or on the Internet. Can you see the similarity between sound waves and the graph of the sine function?

EXERCISES

3 .2

A. Trigonometric Functions and Their Properties

4. Find the sign of each ratio. a. sin 133°

b. cos 224°

c. tan 245°

d. cot 22°

a. sin 0° + 3 ⋅ cos 270° + 4 ⋅ tan 180° – cot 270°

e. tan (–298°)

f. sec 552°

b. 3 ⋅ cos 90° + tan 540° + 2 ⋅ cot 270° – sin 180° 7π 3π π c. sin + 3 ⋅ cos – 2 ⋅ cot 2 2 2 3π 3π 7π d. sec5π – 3 ⋅ csc + 2 ⋅ cot – sin 2 2 2 17 π 11π e. sin7 π + 2 ⋅ cos + 3 ⋅ tan7 π – cot 2 2

g. csc (–467°)

h. sin

1. Evaluate each expression.

8π 7 4π k. cot(– ) 7

i.

cos

21π 5

j.

tan( −

l.

sec

26 π ) 15

23π 5

2. Find the domain of each function. a. f(x) = 2sin(3x – 2) + 1 b. f(x) = 2cos3x – 2

5. In which equations is x undefined?

c. y = 4tan(4x – 2) d. y = 2cot(x + 1) + 3

a. sin x = 0.2

b. cos x = 0.9

e. y = csc(3x – 2)

c. tan x = 3

d. cot x = –0.5

f. y = sec(3x – 60°)

e. sec x = –0.3

f. cos x = 3

g. y = 3sin 2x – 5 cos (7x – 2)

g. csc x = 2

h. tan x = –3

h. y = tan 2x – 3 cot 5x

i. sec x = 34

j. csc x = 0

3. Find the range of each function. a. y = sin 3x

b. y = 5 sin 4x

c. y = 3 sin 2x – 4

d. y = 4cos (x – 4) + 2

k. cos x = 2

e. y = 3 sin (x – 2) + 3 f. y = 2 cos 3x – 2 g. y = 2tan (2x + 1)

6. Find the maximum and minimum values of each

h. y = tan (4x – 2)

function.

i. f(x) = sec 2x + 1

a. y = 3sin x + 4

b. y = 2cos(2x – 1)

j. f(x) = 2 csc 3x – 1

c. y = 5sin 3x – 1

d. y = 2sec x + 1

k. f(x) = tan 2x + cot 3x

e. y = 3csc x

f. y = tan 4x

160

Algebra 10

7. Find the biggest ratio in each set.

11. Given that 0 < x <

π , which of the following 4

a. cos 67°, cos 78°, cos 127°, cos 299°

statements are true?

b. sin 45°, sin 66°, sin 233°, sin 300°

a. sin x < cos x

b. cot x > cos x

c. tan 2°, tan 88°, tan 92°, tan 358°

c. tan x > cot x

d. cot x > sec x

d. sin 70°, cos 50°, tan 40°, cot 30°

e. sin x > cos x

f. sec x > csc x

B. Calculating Trigonometric Values 8. Write each set of ratios in ascending order.

12. Complete the table.

a. x = sin 35°, y = cos 84°, z = tan 255°. b. a = sin 30°, b = cos 230°, c = tan 70°, d = sec 80°. c. m = sin 30°, n = tan 40°, r = cot (–20°). d. x = sin 80°, y = cos 80°, z = tan 80°, k = sec 80°. e. a = tan 20°, b = cot (–40°), c = sec 60°

Trigonometric Values of Quadrantal Angles q in q in sin q Degree Radian (y) 0°

cos q (x)

tan q (y/x)

cot q (x/y)

sec q (1/x)

csc q (1/y)

1 p 2

1 0

180°

0

270° 2p

0

undefined

–1

undefined undefined

undefined undefined

0

0

undefined

–1

1

undefined

13. Evaluate each expression without using a 9. Which of the following statements are true? a. (sin 20° ⋅ sin 40°) < sin 20° b. cos 50° > (cos 50° ⋅ tan 33°) c. (sin 50° ⋅ cos 20°) < sin 50° d. (tan 60° ⋅ cos 40°) < tan 60°

trigonometric table or calculator. a. sin 180° + cos 270° + tan 360° + cot 90° b. sin 90° – cos 270° – (tan 180° ⋅ cot 270°)

14. Find the reference angle for each angle.

e. (tan 250° ⋅ sin 40°) > tan 250°

a. 12°

b. 112°

c. 212°

d. 312°

f. (cos 100° ⋅ sin 80°) > cos 100°

e. 50°

f. 150°

g. 250°

h. 350°

15. Find the reference angle for each angle. π < x < y < π , which of the following 2 statements are true?

10. Given that

a. sin x > cos x

b. tan x > tan y

c. sin x < sin y

d. cos x > cos y

e. sin x > sin y

f. cot x > cot y

Trigonometry: Functions, Equations and Inequalities

a. –25°

b. –140°

c. –245°

d. –305°

e. –5°

f. –95°

g. –260°

h. –320°

16. Find the reference angle for each angle. a. 1000°

b. 3456°

c. –3000°

d. –7890°

e. 2000°

f. 6789°

g. –1000°

h. –2345° 161

17. Find the reference angle for each angle. a. e.

π 11

b.

2π 13

f.

7π 12

c.

9π 15

g.

18 π 13

d.

22π 17

h.

22. Find each trigonometric value by using a reference 25π 14 36π 19

angle. a. c. e. g.

sin (–225°) tan (–300°) sin (–1590°) tan (–9045°)

b. cos (–150°) d. cot (–30°) f. cos (–675°) h. cot (–600°)

18. Find the reference angle for each angle. a. −

π 8

π e. − 6

b. −

7π 10

c. −

8π f. − 9

13π 12

17 π g. − 12

d. −

23π 14

28 π h. − 15

19. Find the reference angle for each angle. 2019 π 9

a.

73π 6

b.

e.

57 π 7

f. 1007 π 73

c. −

101π 13

g. − 602 π 98

d. −

1001π 15

h. − 1009 π 99

Signs of the Trigonometric Values of Nonquadrantal Angles sin θ =

y x y r x r cos θ = tan θ = cot θ = sec θ = csc θ = y y r r x x

I

x>0 y>0

II

x<0 y>0

–

–

III

x<0 y<0

–

+

IV

x>0 y<0

+

+ + +

–

– +

angle. a. sin

5π 6

b. cos

π 3

c. tan

5π 4

d. cot

11π 6

⎛ π⎞ e. sin ⎜ − ⎟ ⎝ 4⎠

⎛ 2π ⎞ f. cos ⎜ − ⎟ ⎝ 3 ⎠

g. tan ⎛⎜ − 5π ⎞⎟ ⎝ 4 ⎠

⎛ 7π ⎞ h. cot ⎜ − ⎟ ⎝ 6 ⎠

24. Find each trigonometric value by using a reference

20. Complete the table with + and – signs. Quadrant, axis

23. Find each trigonometric value by using a reference

–

angle.

67 π 6 55π c. tan 4

100 π 3 607 π d. cot 6

a. sin

b. cos

⎛ 83π ⎞ e. sin ⎜ − ⎟ ⎝ 4 ⎠

⎛ 202 π ⎞ f. cos ⎜ − ⎟ 3 ⎠ ⎝

g. tan ⎛⎜ − 151π ⎞⎟ 4 ⎠ ⎝

⎛ 89π ⎞ h. cot ⎜ − ⎟ ⎝ 6 ⎠

21. Find each trigonometric value by using a reference angle. a. c. e. g. 162

sin 120° tan 315° sin 855° tan 2025°

25. Evaluate the expressions. b. cos 240° d. cot 135° f. cos 3660° h. cot 1410°

a. cos 45° + cos 330° + cos 150° + cos 315° b. cot 5π − sin 5π + tan π − cos 7 π 6 4 3 4 c. sec 300° + tan 585° + cot 765° + csc 1230° Algebra 10

26. Evaluate each expression given that α is an acute angle. a. sin(180° + α) – cos(270° + α) + tan(360° + α) + cot(900° + α) π π b. sin ( π − α )+sin ( π + α )+cot ( − α)+cot ( + α) 2 2

27. For each trigonometric ratio in the given quadrant, find the other trigonometric ratios for θ. 3 a. sin θ = , θ ∈ (0 °, 90 °) 2 b. cos θ = −

2 , θ ∈ (90 °, 180 °) 2

5 c. tan θ = , θ ∈ (180 °, 270 °) 4 2 d. cot θ = − , θ ∈ (270 °, 360 °) 3

28 . For each trigonometric ratio in the given quadrant, find the other trigonometric ratios for θ. 2 ⎛ 3π ⎞ a. sin θ = − , θ ∈ ⎜ , 2 π⎟ 2 ⎝ 2 ⎠ b. cos θ = −

24 3π ⎞ ⎛ , θ ∈ ⎜ π, ⎟ 25 2 ⎠ ⎝

30. Each equation contains a trigonometric function. ›

Find the value of the cofunction of this function in the given quadrant. a. α ∈ (0°, 90°), 3(tan α – 4) = 2tan α – 9 b. α ∈ (90°, 180°), 7(sin α − 1) = c. α ∈ (270°, 360°),

29. α is an acute angle. ›

a. If

3 ⋅ sin α +1 2 = , what is sin α? 4 − 5 ⋅ sin α 5

b. If

cos α − 2 1 = − , what is cos α? 7 ⋅ cos α +11 6

tan α +5 tan α − 4 c. If =− , what is tan α? 6 2 d. (tan 45° ⋅ cot α) + (sec 60° ⋅ cot α) = 12 is

given. Calculate cot α. Trigonometry: Functions, Equations and Inequalities

22+ csc α = –4 3+ 4csc α

31. In the figure,

A

› m(∠ABC) = 150°,

x

AB = 6 and BC = ñ3. Calculate

6

150° B

ñ3

C

b. the area of ΔABC.

a. AC = x.

32. Find the perimeter

D

› P and area A of the

4

C

120°

trapezoid ABCD in the figure.

⎛π ⎞ c. tan θ = −4, θ ∈ ⎜ , π ⎟ ⎝2 ⎠ π⎞ ⎛ d. cot θ = 7, θ ∈ ⎜ 0, ⎟ 2⎠ ⎝

sin α − 10 2

A

B

10

33. Write each angle in degree-minute form. Give your answer rounded to two decimal places in the minutes place. a. 48.5°

b. 136.2°

c. 213.75°

d. 313.79°

34. Write each angle in decimal degrees. Give your answer rounded to two decimal places. a. 121° 15′

b. 346° 50′

c. 198° 19′

d. 23° 56′ 12′′ 163

35. Find each trigonometric value rounded to four decimal places.

40. Find the fundamental period of each function. a. y = sin2 2x – cos2 2x

a. sin 23.4°

b. cos 54.25°

b. y = cos3 5x – 3 cot 2x

c. tan 71.1°

d. cot 63.55°

c. y = cos4 2x – cot 3x + sin5 (3x – 1) d. y = 2 sin x ⋅ cos 4x e. y = 4 cos 5x ⋅ sin 3x

36. Find each trigonometric value rounded to four › decimal places.

f. y = sin x ⋅ sin 3x g. y = cos 3x ⋅ cos 4x

a. sin 121° 15′

h. y = sin2 3x + cos3 3x – tan 2x + cot 3x

b. cos 346° 50′

i. y = tan2 x + cot5 4x – sin2 4x

c. tan 131° 27′

j.

d. cot 89° 49′

C. Periods of Trigonometric Functions

x 3 y = tan x +5 ⋅ sin 4( ) 3

41. Find the fundamental period of y = arctan (tan x).

37. Find the fundamental period of each function. 42. Find the fundamental period of y = sin (πx).

a. y = sin2 3x b. y = 3 sin3 (2 – 5x) c. y = cos4 (x + 3)

D. Graphs of Trigonometric Functions

d. y = 3 sec5 (7 – 3x) – 5

43. Sketch the graph of each function.

e. y = csc12 (5x + 1) – 2 f. y = cos 3x

a. y = 2 sin 3x + 1

b. y = 3 – 2 ⋅ cos x

c. y = 2 tan 3x

d. y = 2cos 2x – 2

e. y = 5sin 4x + 1

38. Find the fundamental period of each function. 44. Draw the graph of each function in the interval

a. y = tan3 (2x – 1)

[0, π].

b. y = 4 tan 5x x c. y = cot 2 (3 – ) 5 d. y = 3 – cot (5x + 1)

a. y = sin 3x b. y = tan 2x – 1 c. y = cot x + 2

x π

39. Find the fundamental period of y = cos( ) . 164

d. y = 2 sin

x 2

e. y = 3cos

x 3 Algebra 10

45. The graph of the function f(x) = cos x is given. Which of the following transformations should be combined to obtain the graph of f(x) = cos (3x + 2) – 1? I.

III. move the graph 1 unit down

V.

47. Write each value in degrees.

move the graph 2 units right move the graph 2 units left

2 ) 2

a. arcsin 0

b. arcsin(–

c. arcsin (–1)

d. arcsin

e. arccos 1

f.

g. arccos 0

h. arccos (–1)

move the graph 1 unit left

II. move the graph 1 unit up

IV.

E. Inverse Trigonometric Functions

2 2

arccos(–

3 ) 2

VI. divide all values on the x–axis by 3

48. Write each value in radians.

VII. multiply all values on the x–axis by 3

46. Find the equation which corresponds to each graph. a. 2

a. arctan 1

b. arctan (–1)

c. arccot 1

d. arccot 0

e. arccot (–ñ3)

f.

arctan(–

3 ) 3

g. arctan (–ñ3)

h. arccot(–

3 ) 3

y

3p/2

–p –p/2 O p/2

p

x

2p

49. Evaluate each expression. a. cos (arcsin 0)

b.

c. sin(arctan

1

e. tan (arccos 1)

y

–p

3 ) 3

–2

y = sin x

p

O

2p

1 b. sin(arccos ) 2

d. cos(arcsin f.

5 ) 13

4 cot(arcsin ) 5

x

–1

c.

y

50. Evaluate each expression.

4

a. tan(arctan

3 2 1 p

–p –2p

O

y = sin x 2p x

3 ) 2

b. sin (arctan 1)

c. cos (arccot ñ3)

d. arccot (tan 30°)

e. arctan (cot 135°)

f. arctan (sin 270°)

–1

Trigonometry: Functions, Equations and Inequalities

165

51. Evaluate each expression. 7 ) 25 3 b. sin(arccos ) 5

54. Find arcsin(sin

15π ). 4

a. cos(arcsin

5 b = arcsin(– )? 6

1 c. cot( arcsin ) 3

d. tan (arccot (–3)) e. cos (arctan 4)

5 6

55. What is a + b if a = arcsin( ) and

56. Find arccos(sin

29π ). 3

f. cot (arccos (sin 60°)) 4 5

2 3

57. Find cos(arcsin – arctan ) . 52. Evaluate each expression. π 4 a. sin( – arccos ) 2 7 π 5 b. tan( + arc cot ) 2 2

c. cot(

58. Find arcsin (cos(arctan(cot 60°))).

59. Given that x ∈ ( π,

3π 1 – arcsin ) 2 2

4 d. tan(2 ⋅ arcsin ) 5

sin x. 3 5

sin(2 ⋅ arcsin

12 ) 13

53. Find the domain of each function.

1 2

60. Find tan(arcsin + arccos ) .

3 e. cot(arccos ) 4

f.

3π 4 ) and x = arctan , find 2 3

61. Find cos (arcsin(tan(arccot 1))).

62. Find tan(arccot(sin(arccos 0)))

a. y = 3 ⋅ arcsin (2x + 4) b. y = 2 ⋅ arccos (3x + 2) c. y = 5 – 2 ⋅ arccos (x – 3)

π 6

63. Find arctan(cos +sin

2π ). 3

d. y = arcsin (3 – 5x) e. y = arctan 3x f. y = arccot (5x + 1) 166

64. Evaluate ›

3 3 3 [arccos + arcsin + arctan 3]. π 2 2 Algebra 10

A. TRIGONOMETRIC THEOREMS We now know how to calculate one or more angles or side lengths in a right triangle from given information about the triangle. However, there are also relations between the angles and sides of any triangle (not just right triangles). In this section we will study these general relations and use them to solve triangle problems.

1. The Law of Cosines law of cosines

Theorem

In a triangle ABC with side lengths a, b and c,

A

a = b + c – 2bc ⋅ cos A 2

2

2

b2 = a2 + c2 – 2ac ⋅ cos B

b

c

c2 = a2 + b2 – 2ab ⋅ cos C.

a

B

Proof In a triangle ABC, a is the side opposite A, b is the side opposite B and c is the side opposite c. A q c

B

a

First we draw CH as an altitude of the triangle. Point H now divides AB into two segments such that AH = x and HB = c – x. Consequently there exist two right triangles: ΔCAH and ΔCHB.

a, b and c can also mean the lengths of sides a, b and c respectively. We also write A to mean the angle at A, B to mean the angle at B and C to mean the angle at C: in the figure, m(A) = θ.

c

x b

c–x

B

We can subtract these side by side:

H

h

(1)

h2 + x2 = b2. (2) C

A

Applying the Pythagorean Theorem to each right triangle gives us h2 + (c – x)2 = a2

b

C

a

C

h2 + (c – x)2 = a2 –––––––––––––––––––––––––––– h2 + x2 = b2 h2 + (c – x)2 – h2 – x2 = a2 –b2 c2 – 2cx + x2 – x2 = a2 – b2. So a2 = b2 + c2 – 2cx.

Trigonometry: Functions, Equations and Inequalities

(3) 167

In the right triangle CAH,

C

x cos A = , i.e. x = b ⋅ cos A. b

Substituting this value of x in (3) gives us a2 = b2 + c2 – 2cb ⋅ cos A.

b

We can rewrite this as a2 = b2 + c2 – 2bc ⋅ cos A, which is the required result.

A

The proofs of the other two identities are similar. They are left as an exercise for you.

h

x

H

Notice that by rewriting the three parts of the law of cosines we obtain the relations cos A =

b2 + c2 – a2 2bc

cos B =

a2 + c2 − b 2 2ac

cos C =

a2 + b 2 – c2 . 2ab

Note By the law of cosines: 1. If two sides of a triangle and the angle between them are known, it is possible to calculate the length of the third side of the triangle. 2. If the three sides of a triangle are known, it is possible to calculate the trigonometric values of the angles in the triangle. EXAMPLE

81

In the figure, ABC is a triangle with

A

AC = 4, BC = 6 and m(∠ACB) = 60°.

4

Find AB. Solution

By the law of cosines, c2 = a2 + b2 – 2ab ⋅ cos C c2 = 62 + 42 – 2 ⋅ 6 ⋅ 4 ⋅ cos 60° 1 c2 = 36 + 16 – 2 ⋅ 6 ⋅ 4 ⋅ 2 c2 = 28

c

60°

C

6 B

c = ò28 = 2ñ7. 168

Algebra 10

EXAMPLE

82

Solution

In the figure, ABC is a triangle with AB = 2, BC = ñ3 and m(∠ABC) = 150°. Find b.

A b 2

150° ñ3

B

By the law of cosines,

C

b2 = a2 + c2 – 2ac ⋅ cos B ⎛ 3⎞ b2 = (ñ3)2 + 22 – 2 ⋅ ñ3 ⋅ 2 ⋅ ⎜ – ⎜ 2 ⎟⎟ ⎝ ⎠ 2 b =3+4+6

⎛ 3⎞ ⎜⎜ cos 150 ° = – ⎟ 2 ⎟⎠ ⎝

b2 = 13 b = ò13.

EXAMPLE

83

The figure shows a triangle ABC with

C

AC = 3, AB = 7 and m(∠ACB) = 120°. Find a. 3

Solution

By the law of cosines, ⎛ 1⎞ 7 2 = a2 + 32 − 2 ⋅ a ⋅ 3 ⋅ ⎜ − ⎟ ⎝ 2⎠

1 (cos 120 ° = − ) 2

49 = a2 +9+ 3 a;

( a – 5)( a+8) = 0;

a2 + 3a − 40 = 0;

a = 5 , a = –8.

(–5)

B

7

A

c2 = a2 + b2 − 2ab ⋅ cos C

a

120°

(8)

Since the length of a side cannot be negative, a = 5.

EXAMPLE

84

Solution

Find cos A in the triangle in the figure.

C

By the law of cosines, b2 + c2 – a2 cos A = 2bc =

52 + 42 – 2 2 2 ⋅5 ⋅4

=

25+16 – 4 40

=

37 . 40

Trigonometry: Functions, Equations and Inequalities

5

A

2

4

B

169

EXAMPLE

85

Solution

Find the measure of angle B in the figure.

B

By the law of cosines, cos B =

2

2

2

2

a +c – b 2ac

=

3 +6 – (3 3) 2 ⋅3⋅6

=

9+ 36 – 27 36

=

18 1 = . 36 2

So cos B =

3

6

2

A 2

3ñ3

C

1 1 . We know that m(∠B) < 90° and cos 60° = . Therefore m(∠B) = 60°. 2 2

Check Yourself 16 1. The side lengths of a triangle are 4, 5 and 6 units respectively. Find the cosine of the smallest angle in the triangle.

Inscribed and circumscribed:

Theorem

4

BC = ò61 and m(∠BAC) = 120°. Find AC.

1. 0.75

120°

B

Answers The black line shows an inscribed triangle. The red line shows a circumscribed circle.

A

2. ABC is a triangle with AB = 4,

x

C

ò61

2. 5

2. The Law of Sines law of sines Let ABC be a triangle with side lengths a, b and c

C

which is inscribed in a circle with radius R. Then

a b c = = = 2 R. sin A sin B sin C

b

A

O

c

a R B OB = R

170

Algebra 10

Proof

Look at the figure.

C

In the right triangle CDA, sin A =

hC , i.e. hC = b ⋅ sin A. b

(1)

In the right triangle CDB, sin B =

hC , i.e. hC = a ⋅ sin B. a

E

F

b

ha

(2) A

From (1) and (2)

hb hc D

B c

we obtain b ⋅ sin A = a ⋅ sin B, i.e. a b = . sin A sin B

a

(I)

Similarly, in the right triangle BFA we have sin A = Finally, in the right triangle BFC, sin C =

hb , i.e. hb = c ⋅ sin A. c

hb which gives hb = a ⋅ sin C. a

From (3) and (4) we obtain c ⋅ sin A = a ⋅ sin C, i.e. From (I) and (II) we obtain

(3)

a b c = = . sin A sin B sin C

(4)

a c = . sin A sin C

(II)

(Result 1)

To show the relation with the radius R of the circumscribed circle, we can use the figure opposite.

A M

M is a point on the circle such that MC is the

O

diameter of the circumscribed circle and m(∠MBC) = 90°. Since ∠BAC and ∠BMC are

B

C

circumscribed angles having the same arc BïC on the circle, m(∠BAC) = m(∠BMC). Since m(∠M) = m(∠A) we have sin M = sin A. In the right triangle MBC, sin M =

So

a a a = , i.e. sin A = . 2R MC 2 R

a = 2 R. sin A

(Result 2)

Combining Result 1 and Result 2 gives us Trigonometry: Functions, Equations and Inequalities

a b c = = = 2 R. sin A sin B sin C 171

Note Let ABC be an inscribed triangle. By the law of sines: 1. If the measures of the three angles and the radius of circumscribed circle are known, it is possible to find the side lengths of the triangle. 2. If the radius of the circumscribed circle and the side lengths of the triangle are known, it is possible to calculate the trigonometric values of the angles of the triangle.

EXAMPLE

86

Solution

In a triangle ABC, m(∠A) = 30° and the length of side a is 8 cm. Find the area of the circumscribed circle of ΔABC. By the law of sines,

A 30°

a 8 = 2 R so = 2 R. sin A sin 30 °

R

8 = 2 R, i.e. R = 8 cm. 1 2 We can calculate the area A of a circle with

So

a=

8

C

B

the formula A = πR2. So the area is π ⋅ 82 = 64π cm2.

EXAMPLE

87

Solution

In a triangle ABC, m(∠B) = 45°, m(∠C) = 30° and the length of side c is 4 cm. Find the length of side b. By the law of sines, 4 b c b = . = , i.e. sin 30 ° sin 45 ° sin C sin B So

172

b 4 = 1 2 2 2

A 4

and b = 4ñ2 cm.

45° B

b 30° C

Algebra 10

EXAMPLE

88

Solution

In a triangle ABC, the measures of the interior angles are 30°, 60° and 90° respectively. The radius of the triangle’s circumscribed circle is 6 cm. Find the perimeter of the triangle ABC. By the law of sines, a b c = = = 2 ⋅ 6 =12. This gives sin 30 ° sin 60 ° sin 90 ° a = sin 30 ° ⋅ 12 =

1 ⋅12 = 6 cm, 2

C

3 ⋅12 = 6 3 cm and 2 c = sin 90° ⋅ 12 = 12 cm.

B

60°

b = sin 60 ° ⋅ 12 =

30°

O

6

A

Since the perimeter P(ΔABC) = a + b + c we can write P(ΔABC) = 6 + 6ñ3 + 12 = 18 + 6ñ3 cm.

Check Yourself 17 1. In a triangle ABC, a = 5 cm and the radius of the circumscribed circle is 5 cm. Find m(∠A). 2. In a triangle ABC, m(∠A) = 30°, m(∠B) = 135° and b = 4ñ2 cm. Find the length of side a. Answers 1. 30° or 150°

2. 4 cm

3. Formulas for the Area of a Triangle Theorem

In a triangle ABC with sides a, b and c, the following properties hold: A( ΔABC ) =

1 ⋅ a ⋅ b ⋅ sin C 2

A( ΔABC ) =

1 ⋅ a ⋅ c ⋅ sin B 2

A( ΔABC ) =

1 ⋅ b ⋅ c ⋅ sin A. 2

Trigonometry: Functions, Equations and Inequalities

A

b

c

B

a

C

173

Proof

Look at the figure. By the general rule for the 1 area of a triangle, A( ΔABC ) = ⋅ b ⋅ hb. (1) 2 Moreover, in the triangle ABH, sin A =

A

hb c

whic gives hb = c ⋅ sin A.

H

c hb

(2)

If we substitute (2) into (1) we obtain 1 A( ΔABC ) = ⋅ b ⋅ c ⋅ sin A which is one of the 2 results to be proven.

b

a

B

C

The other proofs are similar and are left as an exercise for you.

EXAMPLE

89

Solution

In a triangle ABC, A(ΔABC) = 3ñ3 cm2, b = 3 cm and c = 4 cm. Find all the possible measures of angle A. A( ΔABC ) = 3 3=

1 ⋅ b ⋅ c ⋅ sin A 2 1 3 ⋅ 3 ⋅ 4 ⋅ sin A, i.e. sin A = 2 2

We know sin 60 ° = sin 120 ° =

EXAMPLE

90

3 . So m(∠A) = 60° or m(∠A) = 120°. 2

C

In the figure, A(ΔABC) = 5 cm2. Find A(ΔADE).

3 D

Solution

1 A( ΔABC ) = ⋅ AB ⋅ AC ⋅ sin A 2 1 2 ⋅ 5 ⋅ 7 ⋅ sin A, i.e. sin A = 2 7 Since angle A is common to both triangles we 5=

4

A

5

B

2

E

can use the value of sin A for both triangles:

174

A( ΔADE) =

1 ⋅ AD ⋅ AE ⋅ sin A 2

A( ΔADE) =

1 2 ⋅ 4 ⋅ 7 ⋅ = 4 cm 2. 2 7 Algebra 10

Theorem

Heron’s Formula Let ABC be a triangle with sides a, b and c and perimeter a + b + c = 2u. Then A( ΔABC ) = u( u − a )( u − b )( u − c ).

Proof

sin2 A + cos2 A =1 so sin2 A = 1 – cos2 A. By the law of cosines, cos A =

(1)

b2 + c2 − a2 . Substituting this in (1) gives 2bc

2

⎛ b2 + c2 − a2 ⎞ (2bc )2 − ( b2 + c2 − a2 )2 sin A =1 − ⎜ ⎟ = 2bc 4b2 c2 ⎝ ⎠ 2

=

(2 bc − b 2 − c2 + a2 )(2bc + b 2 + c 2 − a 2 ) 4b2 c2

⎡ a2 − ( b2 − 2 bc + c2 ) ⎤⎦ ⎡⎣( b2 + 2 bc + c 2 ) − a 2 ⎤⎦ =⎣ 4b2 c2 ⎡ a2 − ( b − c)2 ⎤⎦ ⎡⎣( b + c)2 − a2 ⎤⎦ =⎣ 4b 2 c2 =

( a + c − b)( a + b − c)( b + c − a)( a + b + c) . 4b2 c2

Let us write each factor of the numerator in terms of u as follows: a + b + c = 2u + –2a = –2a –––––––––––––––––––––––––– b + c – a = 2(u – a)

Now we have sin 2 A =

a + b + c = 2u + –2c = –2c –––––––––––––––––––––––––– a + b – c = 2(u – a)

a + b + c = 2u + –2b = –2b –––––––––––––––––––––––––– a + c – b = 2(u – b).

2( u − a) ⋅ 2( u − c) ⋅ 2( u − b) ⋅ 2 u , i.e. 4b2 c2

=

4u( u − a)( u − b )( u − c) b2 c2

=

2 u( u − a )( u − b )( u − c ). bc

If we use substitute this result in the area equation A( ΔABC ) = A( ΔABC ) =

1 ⋅ b ⋅ c ⋅ sin A we get 2

1 2 ⋅b ⋅c ⋅ u( u − a )( u − b )( u − c ). 2 bc

So A( ΔABC ) = u( u − a )( u − b )( u − c ). Trigonometry: Functions, Equations and Inequalities

175

91

EXAMPLE

Solution

In a triangle ABC, a = 9 cm, b = 10 cm and c = 11 cm. Find A(ΔABC). By Heron’s Formula, let 2u=a + b + c so u =

9+10+11 =15. 2

Then A( ΔABC) = u( u − a)( u − b)( u − c) = 15(15 − 9)(15 −10)(15 −11) = 15 ⋅ 6 ⋅ 5 ⋅ 4 = 30 2 cm 2.

92

EXAMPLE

Solution

In the figure, AB = 6 cm, AC = 10 cm and

A

BC =12 cm. Find AD = x. From Heron’s Formula,

6

12+10+6 2u = a + b + c so u = =14. 2

So A( ΔABC ) = u( u − a)( u − b)( u − c)

B

10

x

D 1555555555552555555555553

C

12

= 14(14 − 12)(14 −10)(14 −6) = 14 ⋅ 2 ⋅ 4 ⋅ 8 = 8 14 cm 2.

But A( ΔABC ) =

Theorem

4 14 1 1 cm. ⋅ AD ⋅ BC, which gives us 8 14 = ⋅ x ⋅ 12, i.e. AD = x = 3 2 2

area of an inscribed triangle Let ABC be a triangle with sides a, b, c which is inscribed in a circle with radius R. Then A( ΔABC ) =

a⋅b⋅c . 4R C

Proof

1 A( ΔABC ) = ⋅ b ⋅ c ⋅ sin A. By the law of sines, 2 a a . = 2 R which gives us sin A = 2R sin A

If we substitute

R for sin A in the equation for 2R

b

A

O

c

a R B OB = R

the area, we obtain A( ΔABC ) = 176

1 a a⋅b⋅c ⋅b ⋅c ⋅ = . 2 2R 4R Algebra 10

EXAMPLE

93

Solution

The side lengths of a triangle are 5 cm, 6 cm and 7 cm. Find the radius of the triangle’s circumscribed circle. C

From Heron’s Formula, 2u=a + b + c so u =

5+6+7 = 9 and 2

R

7

A( ΔABC ) = u( u − a)( u − b )( u − c)

6

O A

= 9(9 − 5)(9 − 6)(9 − 7)

5

B

= 9 ⋅ 4 ⋅ 3 ⋅ 2 = 6 6 cm 2.

However, since A( ΔABC ) = Solving for R gives R =

a⋅b⋅c 5 ⋅ 6 ⋅ 7 210 we have 6 6 = = . 4R 4R 4R

35 cm. 4 6

Check Yourself 18 1. In a triangle ABC, AB = 8 cm, BC = 12 cm and m(∠ABC) = 30°. Find the area of the triangle. 2. A triangle has side lengths of 14 units, 16 units and 20 units respectively. Find its area. 3. In a triangle ABC, A(ΔABC) = 18 cm2 and a ⋅ b ⋅ c = 72. Find the radius of circumscribed circle. Answers 1. 24 cm2

2. 15ò55 square units

3. 1 cm

4. Further Theorems (Optional) Theorem

Let ABC be a triangle with sides a, b and c such that A(ΔABC) = S. Then a=

2S ⋅ sin A , sin B ⋅ sin C

Trigonometry: Functions, Equations and Inequalities

b=

2 S ⋅ sin B , sin A ⋅ sin C

c=

2 S ⋅ sin C . sin A ⋅ sin B 177

EXAMPLE

94

Solution

The interior angles of a triangle ABC measure 30°, 60° and 90° respectively. Given that A(ΔABC) = S = 2ñ3 cm2, find the length of each side. Let us use the theorem. 2⋅2 3 ⋅

2S ⋅ sin A 2 ⋅ 2 3 ⋅ sin 30 ° = = a= sin B ⋅ sin C sin 60 ° ⋅ sin 90 °

3 ⋅1 2 2⋅2 3 ⋅

2S ⋅ sin B 2 ⋅ 2 3 ⋅ sin 60 ° = = b= sin A ⋅ sin C sin 30 ° ⋅sin 90 ° c=

1 ⋅1 2

1 2 = 2 cm

3 2 = 2 3 cm

2S ⋅ sin C 2 ⋅ 2 3 ⋅ sin 90 ° 2 ⋅2 3 ⋅1 = = = 4 cm sin A ⋅ sin B sin 30 ° ⋅sin 60 ° 1 3 ⋅ 2 2

law of tangents

Theorem

In a triangle ABC with sides a, b and c, a+ b = a–b

EXAMPLE

95

Solution

⎛ A+ B ⎞ tan ⎜ ⎟ ⎝ 2 ⎠, ⎛ A – B⎞ tan ⎜ ⎟ ⎝ 2 ⎠

a+c = a–c

⎛ A+C ⎞ tan ⎜ ⎟ ⎝ 2 ⎠, ⎛A–C ⎞ tan ⎜ ⎟ ⎝ 2 ⎠

In the figure, m(∠B) = 105°, m(∠C) = 15° and b + c = 3 +ñ3. Find the lengths of sides b and c.

b+c = b–c

C

15°

⎞ ⎟ ⎠. ⎞ ⎟ ⎠

b

A 105°

By the law of tangents,

b+ c = b–c

⎛ B+ C tan ⎜ ⎝ 2 ⎛B – C tan ⎜ ⎝ 2

c

B

⎛ B+ C ⎞ ⎛ 105 ° +15 ° ⎞ tan ⎜ tan ⎜ ⎟ ⎟ 3+ 3 2 2 ⎝ ⎠ . So ⎝ ⎠ = b–c ⎛ B– C ⎞ ⎛ 105 ° – 15 ° ⎞ tan ⎜ tan ⎜ ⎟ ⎟ 2 ⎝ 2 ⎠ ⎝ ⎠ =

tan (60 °) 3 = tan (45 °) 1

= 3 b – c=

3+ 3 3

=

3 3 3

=

3 3+3 3

= 3 +1. ⎧⎪b + c = 3+ 3 If we solve the system ⎨ , we find b = 2ñ3 and c = 1. ⎪⎩b − c = 3 +1 178

Algebra 10

Check Yourself 19

1 1. In a triangle ABC, a = 8 cm, A(ΔABC) = 16 cm2 and sin B ⋅ sin C = . Find m(∠A). 4

2. In a triangle ABC, b = 11 cm, c = 5 cm, m(∠B) = 78° and m(∠C) = 42°. Find the value of tan 18°. Answers 1. 30° or 150°

2.

3 3 8

B. TRIGONOMETRIC FORMULAS 1. Sum and Difference Formulas In this section we will learn the relations between the sum or difference of two angles and their trigonometric ratios. We will prove these relations using the trigonometric identities we have studied.

a. sin(x±y)

A

In the figure, A(ΔABC) = A(ΔABH) + A(ΔACH). By the formula for the area of a triangle, 1 1 1 ⋅ b ⋅ c ⋅ sin A = ⋅ c ⋅ h ⋅ sin x + ⋅ b ⋅ h ⋅ sin y, 2 2 2 which we can rewrite as (b ⋅ c ⋅ sin A) = (c ⋅ h ⋅ sin x) + (b ⋅ h ⋅ sin y). If we divide both sides by b ⋅ c we get h h sin A = ⋅ sin x + ⋅ sin y. b c

x

y b

c h

B

H

C

a

(1)

In the triangle, m(A) = x + y, cos y = If we substitute these in (1) we obtain

h h and cos x = . c b

sin (x + y) = [sin x ⋅ cos y] + [cos x ⋅ sin y] . If we replace y with –y in this equation we get sin (x + (–y)) = [sin x ⋅ cos (–y)] + [cos x ⋅ sin (–y)]. Since cos (–y) = cos y and sin (–y) = –sin y we have sin (x – y) = [sin x ⋅ cos y] – [cos x ⋅ sin y] EXAMPLE

96

Solution

.

Calculate sin 75°. We can write 75° as the sum or difference of two easier angles.

Trigonometry: Functions, Equations and Inequalities

179

For example, we know the values of the trigonometric functions of 45° and 30°, so we can write 75° = 45°+ 30°. So sin 75° = sin (45° + 30°). By the sine of the sum of two angles, sin(45° + 30°) = (sin 45° ⋅ cos 30°) + (cos 45° ⋅ sin 30°)

EXAMPLE

97

Solution

=

2 3 2 1 ⋅ ⋅ + 2 2 2 2

=

6+ 2 6+ 2 . So sin75° = . 4 4

Calculate sin 15°. We can write 15° as 45° – 30°, 60° – 45° or any other suitable combination. Let us choose 60° – 45°. sin 15° = sin (60° – 45°) = (sin 60° ⋅ cos 45°) – (cos 60° ⋅ sin 45°) =(

3 2 1 2 ⋅ )–( ⋅ ) 2 2 2 2

=

6– 2 4

b. cos(x±y)

A

To find cos (x + y) we will use the formula for the sine of the difference of two angles obtained in the previous section. cos α = sin(90° – α)

x

y b

c

cos (x + y) = sin (90° – (x + y)) = sin (90° – x – y) = sin ((90° – x) – y) (regrouping)

B

C H 155555555525555555553

= sin ((90° – x) – y)

a

= sin [(90° – x) ⋅ cos y] – [cos (90° – x) ⋅ sin y]. Since sin (90° – x) = cos x and cos (90° – x) = sin x, cos (x + y) = [cos x ⋅ cos y] – [sin x ⋅ sin y] . If we replace y with –y in this equation we get cos (x + (–y)) = [cos x ⋅ cos (–y)] – [sin x ⋅ sin (–y)]. Since cos (–y) = cos y and sin (–y) = –sin y we have cos (x –y) = [cos x ⋅ cos y] + [sin x ⋅ sin y] . 180

Algebra 10

EXAMPLE

98

Solution

Calculate cos 105°. We can write 105° as the sum or difference of two easier angles. Let us choose 105° = 60° + 45°. cos 105° = cos (60° + 45°) = (cos 60° ⋅ cos 45°) – (sin 60° ⋅ sin 45°) = =

EXAMPLE

99

Solution 1

1 2 3 2 – ⋅ ⋅ 2 2 2 2 2– 6 4

Show that cos (60° + 30°) ≠ cos 60° + cos 30°. We have 60° + 30° = 90° and we know cos 90° = 0. On the other hand, 1 3 ≠ 0. Therefore, cos (60° + 30°) ≠ cos 60° + cos 30°. cos 60 ° + cos 30 ° = + 2 2

Solution 2

We know that cos (x+ y) = (cos x ⋅ cos y) – (sin x ⋅ sin y). So cos (60° + 30°) = (cos 60° ⋅ cos 30°) – (sin 60° ⋅ sin 30°), i.e. cos (60 ° + 30 °) =

1 3 3 1 ⋅ − ⋅ = 0. 2 2 2 2

However, cos 60 ° + cos 30 ° =

1 3 + ≠ 0. So cos (60° + 30°) ≠ cos 60° + cos 30°. 2 2

c. tan(x±y) We know that tan θ =

sin θ sin ( x + y) (sin x ⋅ cos y)+(cos x ⋅sin y) . So tan ( x+ y) = = by cos θ cos ( x+ y) (cos x ⋅ cos y) – (sin x ⋅sin y)

our previous results. If we divide the numerator and denominator by cos x ⋅ cos y and simplify, we get Trigonometry: Functions, Equations and Inequalities

181

(sin x ⋅ cos y)+(cos x ⋅sin y) cos x ⋅ cos y tan ( x + y) = (cos x ⋅ cos y) – (sin x ⋅sin y) cos x ⋅ cos y sin x ⋅ cos y cos x ⋅ sin y + cos x ⋅ cos y cos x ⋅ cos y = cos x ⋅ cos y sin x ⋅ sin y – cos x ⋅ cos y cos x ⋅ cos y sin x sin y + cos x cos y . This gives us = sin x sin y ⋅ 1− cos x cos y tan ( x+ y) =

tan x+ tan y 1 – (tan x ⋅tan y)

.

If we replace y with –y in this equation we get tan ( x + (– y)) = Since tan (–y) = –tan y we have tan ( x – y) =

EXAMPLE

100 Verify that tan 210° =

Solution

tan x – tan y 1+ (tan x ⋅tan y )

tan x + tan (– y) . 1 – (tan x ⋅ tan (– y))

.

3 . 3

210° = 180° + 30° tan 210 ° = tan (180 °+ 30 °)=

tan 180 °+ tan 30 ° 1 – (tan 180 ° ⋅ tan 30 °)

(by the formula above)

3 3 = 3 = 3 3 1– 0 ⋅ 3 0+

EXAMPLE

101 Find tan (x – y) using tan x = 52

Solution

5 1 – tan x – tan y tan ( x − y) = = 2 4 = 1+(tan x ⋅ tan y) 1+ 5 ⋅ 1 2 4 =

182

1 and tan y = . 4 9 4 13 8

18 13 Algebra 10

d. cot(x±y) cot ( x + y) =

1 cos ( x + y) (cos x ⋅ cos y) – (sin x ⋅sin y) = = tan( x + y) sin ( x + y) (sin x ⋅ cos y)+(cos x ⋅sin y)

Let us divide the numerator and denominator by sin x ⋅ sin y and simplify: (cos x ⋅ cos y) – (sin x ⋅sin y) sin x ⋅ sin y cot ( x + y) = (sin x ⋅ cos y)+(cos x ⋅sin y) sin x ⋅ sin y cos x ⋅ cos y sin x ⋅ sin y – sin x ⋅ sin y sin x ⋅ sin y = sin x ⋅ cos y cos x ⋅ sin y + sin x ⋅ sin y sin x ⋅ sin y cos x cos y ⋅ –1 sin x sin y . So = cos y cos x + sin y sin x cot ( x+ y) =

(cot x ⋅ cot y) – 1 cot y+ cot x

If we replace y with –y in this equation we get cot ( x + ( − y)) = Since cot (–y) = – cot y we have cot ( x − y) =

cot ( x – y) =

Note

. cot x ⋅ cot ( − y) – 1 . cot ( − y)+ cot x

–(cot x ⋅ cot y) – 1 , i.e. – cot y + cot x (cot x ⋅ cot y) +1 cot y – cot x

.

We can also calculate these results by using the corresponding results for the tangent and the 1 1 ). fact that cot α = (so cot ( x ± y) = tan ( x ± y) tan α

EXAMPLE

102 Calculate cot 75°.

Solution

cos 75° = cot(45° + 30°) = =

(cot 45° ⋅ cot 30°) – 1 (1 ⋅ 3) – 1 3–1 ( 3 – 1)( 3 – 1) = = = cot 45° + cot 30° 1+ 3 1+ 3 ( 3 +1)( 3 – 1) ( 3 – 1)2 ( 3)2 – ( 1)2

Trigonometry: Functions, Equations and Inequalities

=

3 – 2 3+1 4 – 2 3 = =2– 3 2 2 183

Check Yourself 20 1. Calculate cos 15° and sin 105°. 2. Calculate tan 195° and cot 285°. 3. Verify the results. 1 2 b. cos 300° = 2 2 4. Calculate tan 15° + cot 15°. 4 5. cot x = –1 and cot y = – are given. Find cot (x – y). 3 Answers 2+ 6 1. cos 15° = sin 105° = 2. tan 195° = 2 – ñ3, cot 285° = ñ3– 2 4

a. sin 135° =

4. 4

5. –7

2. Double-Angle and Half-Angle Formulas We now know formulas to calculate trigonometric ratios such as sin(x + y), cos(x + y), tan(x + y) and cot(x + y). In this section we will consider the special case x = y and find formulas for the trigonometric ratios sin 2x, cos 2x, tan 2x and cot 2x. These formulas are called the double-aangle formulas.

a. sin 2x We know that sin (x + y) = (sin x ⋅ cos y) + (cos x ⋅ sin y). If x = y this formula becomes sin (x + x) = (sin x ⋅ cos x) + (cos x ⋅ sin x), i.e. sin 2 x = 2sin x ⋅ cos x We can also rewrite this as sin x ⋅ cos x =

EXAMPLE

sin 2 x . 2

103 Calculate sin 120° using the double-angle formula for sine.

Solution

sin 120° = sin (2 ⋅ 60°) = 2 ⋅ sin 60° ⋅ cos 60° =2⋅ =

184

.

(double-angle formula)

3 1 ⋅ 2 2

3 2

Algebra 10

EXAMPLE

104 Evaluate the expressions.

a. sin 22.5° ⋅ cos 22.5° ⋅ cos 45°

Solution

a. Using sin x ⋅ cos x =

b. 6 sin

π π cos 8 8

sin 2 x gives us 2 1 sin (2 ⋅ 22.5°) ⋅ cos 45° 2 1 = sin 45° ⋅ cos 45° 2

sin 22.5° ⋅ cos 22.5° ⋅ cos 45° =

b. 6sin

1 1 ⋅ sin (2 ⋅ 45°) 2 2

=

1 1 sin 90° = . 4 4

2 π π π π π π cos = 3 ⋅ (2sin cos ) = 3 ⋅ sin(2 ⋅ ) = 3 ⋅ sin = 3⋅ 8 8 8 8 8 4 2

=

EXAMPLE

=

3 2 2

6 x sin 6 x – . 105 Simplify cos cos 2 x sin 2 x

Solution

cos 6 x sin 6 x (cos 6 x ⋅ sin 2 x) – (cos 2 x ⋅sin 6 x) (sin 2 x ⋅ cos6x) – (sin6 x ⋅cos 2 x) – = = . cos 2 x sin 2 x cos 2 x ⋅ sin 2 x sin 2 x ⋅cos 2 x

Using the double-angle formula gives us double-angle formula again gives us

EXAMPLE

106 Simplify

Solution

sin(2 x – 6 x) sin (–4 x) = , and using the sin 2 x ⋅ cos 2 x cos 2 x ⋅ sin 2 x

– sin 4 x = –2. 1 sin 4 x 2

4 ⋅ cos 50 ° ⋅ sin 50 ° ⋅ cos 100 ° . sin 200 °

We can rewrite this as

2 ⋅ (2 ⋅ cos 50 ° ⋅ sin 50 °) ⋅cos 100 ° and use sin 100° = 2 ⋅ sin 50° ⋅ cos 50°. sin 200 °

2 ⋅ sin 100 ° ⋅ cos 100 ° by the double-angle formula. sin 200 ° sin 200 ° Using the double-angle formula again gives us =1 . sin 200 °

This gives us

Trigonometry: Functions, Equations and Inequalities

185

EXAMPLE

107 Evaluate sin 20° ⋅ cos 40° ⋅ cos 80°.

Solution

sin 20 ° ⋅ cos 40 ° ⋅ cos 80 °=

(sin 20 ° ⋅ cos 20 °) ⋅ cos 40 ° ⋅cos 80 ° cos 20 °

1 (sin 40 ° ⋅ cos 40 °) ⋅ cos 80 ° =2 cos 20 °

sin 160° = sin 20° sin a

1 sin 80 ° ⋅ cos 80 ° =4 = cos 20 °

160° 20°

1 sin 160 ° 8 = cos 20 °

1 sin 20 ° 1 8 = tan 20 ° cos 20 ° 8

b. cos 2x We know that cos (x + y) = cos x ⋅ cos y – sin x ⋅ sin y. If x = y this formula becomes cos (x + x) = (cos x ⋅ cos x) – (sin x ⋅ sin x), i.e. cos 2 x = cos2 x – sin2 x

.

We can also use the identities sin2 x = 1 – cos2 x and cos2 x = 1 – sin2 x to obtain two additional formulas: cos 2 x = 2cos2 x – 1 cos 2 x = 1 – 2sin2 x

EXAMPLE

108 Calculate cos 2x given cos x =

Solution

cos 2 x = 2 ⋅ cos 2 x – 1= 2 ⋅( =–

EXAMPLE

.

7 π and 0 < x < . 4 2

7 2 7 −1 ) −1= 2 ⋅ 4 16

1 8

1 – cos 2 x

109 Simplify the expression 1+ cos 2 x .

Solution

1 – cos 2 x 1 – (1 – 2 sin 2 x) 1 – 1+ 2 sin 2 x 2 sin 2 x = = = 1+ cos 2 x 1+(2 cos 2 x – 1) 1+ 2 cos 2 x – 1 2 cos 2 x = tan 2 x

186

Algebra 10

EXAMPLE

110 Given cos 11° = t, write sin 68° in terms of t.

Solution

sin 68° = cos 22°

(cofunctions)

cos 22° = cos (2 ⋅ 11°) = 2 ⋅ cos 11° – 1 2

(double-angle formula)

2

= sin 68° = 2t – 1

c. tan 2x We know that tan ( x + y) =

tan x + tan y . 1 – tan x ⋅ tan y

If x = y this formula becomes tan ( x + x) =

tan x + tan x , i.e. 1 – tan x ⋅ tan x tan 2 x =

EXAMPLE

111

Solution

2tan x 1 – tan 2 x

.

4 tan x = . Find tan 2x. 3 tan 2 x =

2 tan x 1 – tan 2 x

4 8 8 3 = 3 = 3 = 2 16 7 ⎛4⎞ 1– – 1– ⎜ ⎟ 9 9 ⎝3⎠ 2⋅

=–

24 7

d. cot 2x We know that cot ( x + y) =

cot x ⋅ cot y – 1 . cot x + cot y

If x = y this formula becomes cot ( x + x) =

cot x ⋅ cot x – 1 , i.e. cot x + cot x cot 2 x =

Trigonometry: Functions, Equations and Inequalities

cot 2 x – 1 2cot x

. 187

EXAMPLE

112 Given that x is an acute angle, calculate cot 2x using tan x – cot x = 2.

Solution

Since tan x =

1 1 we have – cot x = 2, i.e. cot x cot x

1 – cot 2 x 2 = . cot x 1

This gives us 1 – cot2 x = 2cot x, i.e. cot2 x + 2cot x – 1 = 0. If we apply the quadratic formula we get cot x =

–2+ 6 –2 – 6 . or cot x = 2 2

Since x is an acute angle, the cotangent value must be positive. Since ñ6 > 2, –2 + ñ6 is positive and so cot x =

–2+ 6 . 2

e. Half-angle formulas We have just seen that sin 2x, cos 2x, tan 2x, and cot 2x can be expressed in terms of sin x, cos x, tan x and cot x respectively. In addition, we can apply the procedure in reverse order to express sin x, cos x, tan x, and cot x in terms of sin 2x, cos 2x, tan 2x and cot 2x respectively. For this reason, the double-angle formulas are also called half-angle formulas. By using the double-angle formula for the cosine function, we can obtain the half-angle formulas for the sine, tangent and cotangent functions as follows. x We know that cos 2x = 2 cos2 x – 1. If we replace x with , then 2 x ⎛ x⎞ cos ⎜ 2 ⋅ ⎟ = 2 cos 2 – 1, 2 ⎝ 2⎠ cos x = 2 cos 2

1+ cos x x x – 1 i.e. cos = ± . 2 2 2

(1)

Similarly, cos 2x = 1 – 2sin2 x. If we replace x with cos x = 1 − sin 2

1 − cos x x x i.e. sin = ± . 2 2 2

x x ⎛ x⎞ then cos ⎜ 2 ⋅ ⎟ =1 – sin 2 , 2 2 ⎝ 2⎠

(2)

Using (1) and (2) we can write 1 – cos x x ± sin x 2 2= tan = , i.e. tan x = ± 1 − cos x . 2 cos x 1+ cos x 2 1+ cos x ± 2 2 188

Algebra 10

Similarly, 1+ cos x x ± cos x 2 2= cot = , i.e. 2 sin x 1 – cos x ± 2 2

cot

EXAMPLE

113 cos 2 x = – 51

1+ cos x x =± 2 1 – cos x

.

is given. Find cos x if x is in the first

quadrant.

Solution

cos 2x = 2cos2 x – 1 so cos x = ±

1 + cos 2 x . 2

Since x is in the first quadrant, the cosine is positive. ⎛ 1⎞ 1+ ⎜ − ⎟ ⎝ 5⎠ So cos x = 2 =

EXAMPLE

2 . 5

114 Calculate sin 22.5º using half-angle formulas.

Solution

cos 2x = 1 – 2sin2 x so sin x = ±

So sin (22.5 °) = = Trigonometry: Functions, Equations and Inequalities

2 2

1– 2

1 – cos 2 x . Since x is an acute angle, the sine is positive. 2

(cos 45° =

2 ) 2

2− 2 . 2 189

EXAMPLE

115

Solution

cos x =

3 is given. Find tan x if x is in the fourth quadrant. 2 5

Since x is in the fourth quadrant, 270° < x < 360°. x x is in the second quadrant, where the tangent function < 180 °. This means that 2 2 is negative.

So 135° <

EXAMPLE

By the half-angle formula, tan

1 – cos x x =± and we take the negative value. 2 1+ cos x

3 1– x 5 Hence, tan = – =– 3 2 1+ 5

2 5 , i.e. tan x = – 1 = – 1 . 8 2 4 2 5

116 Find the values of sin 105° and cos 15° using half-angle formulas.

Solution

sin 105 ° = sin

210 ° 1 – cos 210 ° = 2 2

(cos 210 °= –

3 ) 2

Notice that we take the positive value in the half-angle formula because the sine function is positive in the second quadrant. So we have

sin 105 ° =

3 ) 2 = 2+ 3 = 2+ 3 . 2 4 2

1 – (–

Similarly, cos 15 ° = cos

30° 1+ cos 30 ° = . 2 2

(cos 30 °=

3 ) 2

We take the positive value because the cosine function is positive in the first quadrant. So

cos 15 ° = = 190

3 2 = 2+ 3 2 4

1+

2+ 3 . 2 Algebra 10

EXAMPLE

117 Given cot 2x = t, find sin x in terms of t.

Solution

We can show cot shown opposite.

x t = in a right triangle, as 2 1

t2 + 1

1

From the double-angle formula for the sine we x x have sin x = 2 ⋅ sin ⋅ cos . 2 2 1 t ⋅ So sin x = 2 ⋅ 2 2 t +1 t +1 2t = 2 . t +1 EXAMPLE

118 Given

sin x =

angle. Solution

x 2

t

3 x , find cos if x is an acute 5 2

We can solve this problem using the previous formulas. However, let us look at an alternative geometric solution.

C x 2 5 x 2 D

3

x A

B

Step 1: We sketch the triangle ABC as in the figure. We calculate AB = 4 using the Pythagorean Theorem. Step 2: We extend side AB to create DB such that AD = AC = 5. Step 3: We construct the right triangle DBC. Since AD = AC, the triangle DAC is an isosceles triangle. Hence, m(∠CDA) = m(∠ACD) and m(∠CAB) = m(∠CDA) + m(∠ACD). x Therefore, m( ∠CDB) = . Moreover, BD = AD + AB = 5 + 4 = 9. 2

By using the Pythagorean Theorem with the triangle CDB, DC 2 = BC 2 + BD 2 = 32 +9 2 DC = 9+81 = 90 = 3 10. So cos

9 x DB = = 2 DC 3 10 =

3 10 . 10

Trigonometry: Functions, Equations and Inequalities

191

Check Yourself 21 1. α is an acute angle such that sin

α 3 = . Find sin α. 2 5

2. α is an acute angle such that cos α = 3. cos 2 x ⋅ cos x ⋅ cos

2 . Find cos 2α. 5

x x 1 is given. Find cos 8x. ⋅ sin = 2 2 24

4. If tan 84° = t, find cot 78° in terms of t. 5. Calculate tan 15°. Answers 1.

24 25

2. –

17 25

3. 7 9

4.

2t t2 – 1

5. 2–ñ3

4. Sum to Product Formulas In this section we will study formulas which allow us to rewrite the sum of two trigonometric functions as a product of two trigonometric functions. We will use the sum and difference formulas in our working.

a. sin a ± sin b To obtain a formula for sin a + sin b we add the formulas for the sine of a sum and the sine of a difference: sin ( x+ y) =

(sin x ⋅ cos y)+(co s x ⋅ sin y)

+sin ( x − y) =+ (sin x ⋅ cos y) – (cos x ⋅ sin y) sin ( x + y)+ sin ( x − y) = 2 ⋅ sin x ⋅ cos y.

(1)

⎧⎪ a = x + y If we write a = x + y and b = x – y we obtain the system ⎨ . ⎪⎩ b = x – y

Solving this system for x and y gives x =

a+ b a–b and y = , which we can substitute in (1): 2 2

sin a +sin b= 2 sin 192

a+b a –b . cos 2 2

(2) Algebra 10

If we subtract the two formulas instead of adding them, we get sin ( x+ y) = (sin x ⋅cos y)+(co s x ⋅ sin y) – sin ( x – y) = – (sin x ⋅cos y) – (cos x ⋅s in y) sin ( x + y) – sin ( x – y) = 2 ⋅ cos x ⋅sin y.

(3)

If we write a = x + y and b = x – y we obtain the system ⎧⎪ a = x + y a+ b a–b and y = , which we can substitute in (3) : . Solving this gives x = ⎨ 2 2 ⎪⎩ b = x – y sin a – sin b= 2 cos

EXAMPLE

a+b a –b sin 2 2 .

(4)

119 Evaluate the expressions. a. sin 105° + sin 15° b. sin 70° – sin 10°

Solution

We could evaluate each term separately and then combine them. Howewer, the sum to product formulas give us an easier way. ⎛ 105 ° +15 ° ⎞ ⎛ 105 ° – 15 ° ⎞ a. sin 105 ° + sin 15 ° = 2 ⋅ sin ⎜ ⎟ ⋅cos ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠ = 2 ⋅ sin 60 ° ⋅ cos 45 ° 3 2 ⋅ 2 2

=2⋅ =

b.

6 2

⎛ 70° – 10 ⎞ ⎛ 70 °+10 ° ⎞ sin 70 ° – sin 10 ° = 2 ⋅ cos ⎜ ⎟ ⋅sin ⎜ ⎟ 2 2 ⎝ ⎠ ⎝ ⎠ = 2 ⋅ cos 30 ° ⋅ sin 40 ° =2⋅

3 ⋅ sin 40 ° 2

= 3 ⋅ sin 40 °

Trigonometry: Functions, Equations and Inequalities

193

b. cos a ± cos b Now let us add the formulas for the cosine of a sum and the cosine of a difference: cos (x + y) = + cos (x – y)

(cos x ⋅ cos y) – (sin x ⋅ sin y)

= +(cos x ⋅ cos y) + (sin x ⋅ sin y)

cos (x + y) + cos (x – y) = 2 ⋅ cos x ⋅ cos y.

(5)

If we write a = x + y and b = x – y then we obtain the system ⎧⎪ a = x + y a+ b a–b and y = , which we can substitute in (5): . Solving this gives x = ⎨ 2 2 ⎪⎩ b = x – y a+b a –b . cos 2 2 If we subtract the formulas instead of adding them, we get cos ( x+ y) = ( cos x ⋅ cos y) − (sin x ⋅ sin y) cos a +cos b= 2 cos

(6)

− cos ( x − y) = − ( cos x ⋅ cos y)+(si n x ⋅ sin y) cos ( x + y) − cos ( x − y) = −2 ⋅ sin x ⋅ sin y.

(7)

If we write a = x + y and b = x – y we obtain the system ⎧⎪ a = x + y a+ b a–b and y = which we can substitute in (7): . Solving gives x = ⎨ 2 2 ⎪⎩ b = x – y cos a – cos b= – 2 sin

EXAMPLE

120 Simplify

Solution

a+b a –b sin . 2 2

(8)

cos 105 ° + cos 15 ° . sin 105 ° + sin 15 °

cos 105 ° + cos 15 ° = sin 105 ° + sin 15 ° =

105 ° +15 ° 105 ° −15 ° ⋅ cos 2 2 105° +15 ° 105 ° −15 ° ⋅ cos 2 sin 2 2

2 cos

(sum to product formulas)

cos 60 ° ⋅ cos 45 ° sin 60 ° ⋅ cos 45 °

1 1 2 = = . 3 3 2 194

Algebra 10

EXAMPLE

121 Evaluate

Solution

π sin a – sin b given that a + b = . 3 cos a – cos b

a+b a–b 2 cos sin sin a – sin b 2 2 = cos a – cos b –2 sin a + b sin a – b 2 2 a+ b 2 = – cot a + b = a+b 2 – sin 2 cos

π a+ b = – cot = – cot 3 2 2 = – cot

EXAMPLE

π =– 3 6

122 Find the simplest form of the expression

Solution

sin 3 x + sin 5 x+ sin 7 x . cos 3 x + cos 5 x+ cos 7 x

We can use the first and last terms of both the numerator and the denominator in the sum to product formulas. 7 x + 3x 7 x – 3x 2 sin cos + sin 5 x sin 3 x + sin 5 x+ sin 7 x 2 2 = cos 3 x + cos 5 x+ cos 7 x 2 cos 7 x + 3x cos 7 x – 3x + cos 5 x 2 2

Trigonometry: Functions, Equations and Inequalities

=

2 sin 5 x ⋅ cos 2 x+ sin 5 x 2 cos 5 x ⋅ cos 2 x+ cos 5 x

=

sin 5 x ⋅ (2 cos 2 x+1) cos 5 x ⋅ (2 cos 2 x+1)

=

sin 5 x = tan 5 x cos 5 x 195

c. tan a ± tan b For two real numbers a and b not equal to kπ +

π (k ∈ ] ), we know that 2

tan a + tan b =

sin a sin b + . Multiplying gives us cos a cos b

tan a + tan b =

(sin a ⋅ cos b)+(cos a ⋅sin b) , i.e. cos a ⋅ cos b

sin ( a + b) . cos a ⋅ cos b

tan a +tan b=

(9)

If we write – b instead of b in this equation, we get tan a + tan (– b) =

sin ( a + (– b)) , i.e. cos a ⋅ cos (– b) tan a – tan b=

EXAMPLE

(10)

123 Evaluate tan 75° + tan 15°.

Solution

We can use the sum to product formula for the tangent. tan 75 ° + tan 15 ° =

sin (75 ° +15 °) cos 75 ° ⋅ cos 15 °

=

sin (90 °) sin 15 ° ⋅ cos 15 °

=

=

196

sin ( a – b) . cos a ⋅ cos b

1 sin 30 ° 2 1 =4 1 1 ⋅ 2 2 Algebra 10

EXAMPLE

124 Given sin 10° = t, evaluate tan 70° + tan 10° in terms of t.

Solution

tan 70 ° + tan 10 ° =

sin (70 ° + 10 °) cos 70 ° ⋅ cos 10 °

=

sin (80 °) cos 70 ° ⋅ cos 10 °

=

cos 10 ° 1 = sin 20 ° ⋅ cos 10 ° sin 20 °

=

1 1 = 2 sin 10 ° ⋅ cos 10 ° 2 ⋅ t ⋅ 1 – t 2

d. cot a ± cot b For two real numbers a and b not equal to kπ + cot a + cot b =

π (k ∈ ]), we know that 2

cos a cos b + . sin a sin b

Multiplying gives us cot a+ cot b=

cos a ⋅ sin b + cos b ⋅ sin a , i.e. sin a ⋅ sin b

cot a +cot b=

sin ( a + b) sin a ⋅ sin b

.

(11)

sin ( a – b) . sin a ⋅ sin b

(12)

If we write –b instead of b in this equation, we get cot a + cot (– b) =

sin ( a +(– b)) , i.e. sin a ⋅ sin (– b) cot a – cot b= –

EXAMPLE

125 18x = π is given. Find the simplest form of the expression

Solution

cot 9 x + cot x . cot 9 x – cot x

sin (9 x + x) cot 9 x + cot x sin 9 x ⋅ sin x = – sin 10 x = sin (9 x – x) cot 9 x – cot x sin 8 x – sin 9 x ⋅ sin x =–

sin(18 x – 8 x) sin8 x

=–

sin ( π − 8 x) sin 8 x

=–

sin 8 x = –1 sin 8 x

Trigonometry: Functions, Equations and Inequalities

197

Check Yourself 22 Simplify each expression. 1. sin 75° – sin 15° 2. cos 105° – cos 75° 3. tan 165° + tan 15° 4. cot 75° – cot 195° Answers 2 1. 2. –2sin 15° = 2

2– 6 2

3. 0

4. –2ñ3

5. Product to Sum Formulas If we need to convert a product of two trigonometric expressions to a sum or difference we use the product to sum formulas. These formulas are derived from the sum and difference formulas.

a. sin x ⋅ sin y For any real numbers x and y, cos ( x + y) = (cos x ⋅ cos y) – (sin x ⋅ sin y) − cos ( x − y) = –(cos x ⋅ cos y)+(sin x ⋅ sin y) cos ( x + y) – cos ( x − y) = –2 ⋅ sin x ⋅ sin y

sin x ⋅ sin y = –

EXAMPLE

1 . y) – cos (x – y)] . [cos (x + 2

(13)

126 Write sin 5α ⋅ sin 3α as the difference of two trigonometric ratios.

Solution

Let us use the product to sum formula for sine: 1 sin 5α ⋅ sin 3α = – [cos (5 α + 3 α) – cos (5 α – 3 α)] 2 1 = – [cos (8α ) – cos (2 α)]. 2

198

Algebra 10

EXAMPLE

127 Evaluate sin 75°⋅ sin 15°.

Solution

We could evaluate the two terms separately. However, let us use the product to sum formula for sine: sin 75 ° ⋅ sin 15 ° = –

1 [cos (75° + 15°) – cos (75° – 15°)] 2

1 = – [cos 90 ° – cos 60 °] 2 =–

1⎡ 1⎤ 1 0 – ⎥= . ⎢ 2⎣ 2⎦ 4

b. cos x ⋅ cos y For any real numbers x and y, cos ( x+ y) =

(cos x ⋅ cos y) – ( sin x ⋅ sin y)

+cos ( x – y) =+ ( cos x ⋅ cos y)+(sin x ⋅ sin y) cos ( x + y)+ cos ( x – y) = 2 ⋅ cos x ⋅ cos y

cos x ⋅ cos y =

EXAMPLE

1 [cos (x + y) + cos (x – y)] . 2

(14)

128 Write cos 7α . cos 4α as the sum of two trigonometric ratios.

Solution

By the product to sum formula for cosine, cos 7α ⋅ cos 4α =

1 [cos (7α + 4α) + cos (7α – 4α)] 2

1 = [cos (11α )+ cos (3 α)]. 2 EXAMPLE

129 Evaluate cos 105° ⋅ cos 15°.

Solution

cos 105° ⋅ cos 15 ° =

1 [cos (105° + 15°) + cos (105° – 15°)] 2

1 = [cos 120 °+ cos 90 ° ] 2 = Trigonometry: Functions, Equations and Inequalities

1⎡ 1 1 ⎤ – +0 ⎥ = – 2 ⎢⎣ 2 4 ⎦ 199

c. sin x ⋅ cos y For any real numbers x and y, sin ( x+ y) =

(sin x ⋅ cos y)+( cos x ⋅ sin y)

+ sin ( x − y) =+ ( sin x ⋅ cos y) – (cos x ⋅ s in y ) sin ( x + y)+ sin ( x − y) = 2 ⋅ sin x ⋅ cos y

sin x ⋅ cos y =

EXAMPLE

3π

1 [sin (x + y) + sin (x – y)] . 2

(15)

π

130 Evaluate sin 8 ⋅ cos 8 .

Solution

By the product to sum formula, sin

3π π 1⎡ ⎛ 3π π ⎞ ⎛ 3π π ⎞⎤ ⋅ cos = ⎢sin ⎜ + ⎟+ sin ⎜ – ⎟ ⎥ 8 8 2⎣ ⎝ 8 8⎠ ⎝ 8 8 ⎠⎦ =

1⎡ π π⎤ sin + sin ⎥ ⎢ 2⎣ 2 4⎦

=

1⎡ 2 ⎤ 2+ 2 . ⎢1+ ⎥= 2⎣ 2 ⎦ 4

d. cos x ⋅ sin y For any real numbers x and y, sin ( x+ y) =

( sin x ⋅ cos y)+(co s x ⋅ sin y )

– sin ( x – y) = – ( sin x ⋅ cos y) – (cos x ⋅ sin y) sin ( x + y) – sin ( x – y) = 2 ⋅ cos x ⋅ sin y

cos x ⋅ sin y = 200

1 [sin (x + y) – sin (x – y)] 2

.

(16) Algebra 10

EXAMPLE

131 Evaluate cos 8π ⋅ sin 38π.

Solution

cos

π 3π 1 ⎡ ⎛ π 3π ⎞ ⋅ sin = ⎢sin ⎜ + ⎟ – sin 8 8 2⎣ ⎝8 8 ⎠

⎛ π 3π ⎞⎤ ⎜ – ⎟⎥ ⎝ 8 8 ⎠⎦

=

1⎡ π π π⎤ ⎛ π ⎞⎤ 1 ⎡ sin – sin ⎜ – ⎟ ⎥ = ⎢sin + sin ⎥ 2 ⎢⎣ 2 2 4⎦ ⎝ 4 ⎠⎦ 2 ⎣

=

1⎡ 2 ⎤ 2+ 2 ⎢1+ ⎥= 2⎣ 2 ⎦ 4

Note Example 101 and Example 102 are actually the same question. The cosine and sine functions were simply in a different order, so the result does not change due to the commutative property of multiplication. Although we applied a different formula in each question, the results are the same. Therefore we can say that formulas (15) and (16) are different forms of the same identity.

EXAMPLE

132 Evaluate cos 15° ⋅ sin 75°.

Solution

cos 15 ° ⋅ sin 75 ° =

1 [sin(15° + 75°) – sin(15° – 75°)] 2

1 1 = [sin 90 ° – sin (–60 ° )]= [sin 90° + sin 60°] 2 2 =

(sin(–60°) = –sin 60°)

1⎡ 3 ⎤ 2+ 3 ⎢1+ ⎥= 2⎣ 2 ⎦ 4

Check Yourself 23 Evaluate each expression. 1. sin 105° ⋅ sin 15° 2. cos 75° ⋅ cos 15° 3. sin 75° ⋅ cos 15° Answers 1 1 1. 2. 4 4

3.

2+ 3 4

Trigonometry: Functions, Equations and Inequalities

201

APPLICATIONS OF TRIGONOMETRY Trigonometry has many applications in the modern world, in fields as diverse as engineering, architectural design, acoustics, astronomy, physics, biology, map-making, computer graphics, optics and statistics. People working in these fields often use trigonometry without knowing it, because the math is hidden in modern instruments and computer programs. However, the trigonometric formulas are still there, and without these formulas, science and technology would be very different. Here are just three examples of how trigonometry applies to the real world. Trigonometry and Optics Light travels at different speeds through different objects, and we can use trigonometry to calculate how fast light will travel through a transparent object such as air, water or glass. The relative speed of light is controlled by a number called the refraction index: the refraction index of air is very close to 1, the index for water is 1.33, and the index for glass is 1.5. The figure at the right shows a pyramid with light passing through it at an angle θ. The θ α sin( + ) 2 2 . refraction index n for this pyramid is given by n = θ sin 2

a

q

For example, if α = 60° then the refraction index is θ α θ θ θ 3 θ 1 sin( + ) sin ⋅ cos 30 °+cos ⋅sin30 ° sin ⋅ +cos ⋅ 2 2 2 2 2 2 2 2 = 3 + 1 cot θ . n= = = θ θ θ 2 2 2 sin sin sin 2 2 2

Trigonometry and the Climate If you measure the temperature of a pan of water at intervals as you heat it up and leave it to cool, you will find that the temperatures do not rise and fall in a straight line. In fact, the changes in temperature follow a curve defined by y = A ⋅ cos a(t + b) + B. In this equation, A is called the amplitude of the function, a is the period, b is called the phase difference and B is a constant number. As an example, imagine you want to show a graph of the temperature T in a city during a particular period. At 15:00 the temperature is 30° C and at 03:00 the next morning it is 10° C. Assuming that the temperature always falls, how can we draw its graph? We calculate the amplitude A as the average of the maximum and minimum values of the function: 2π π 30° C − 10 ° C = , and so the equation for the temperature is =10. The period is a = A= 24 12 2 π T = A ⋅ cos a(t + b) + B =10 ⋅ cos ( t + b)+ B. 12 t is the time. If we say t = 3 at 03:00 and t = 15 at 15:00 then we have T = 10° C at t = 3 and T = 30° C at t = 15. Using these values, we can find b = 9 and B = 20.

In conclusion, the graph of the temperature in the city is given by π T =10 cos ( t +9)+ 20. 12

North Pole Plane of a parallel

How can we measure the shortest distance between any two points on Earth? This question is important for airline companies and ships all over the world, and we can use trigonometry to find the answer. Navigators divide the Earth using lines of latitude and longitude. To

Equator

Prime Meridian

Half of a great circle

Trigonometry and Navigation

Plane of equator

Plane of a parallel

understand what this means, imagine the intersection of the earth with a South Pole

plane passing through the North and South Poles. The intersection is a complete circle, and half of this circle (from pole to pole) is called a meridian of longitude, or simply a meridian. The meridian which passes through the city of Greenwich in England is called the prime meridian. meridian. This plane also creates a circle. The biggest such circle around the Earth is called the equator. To find the shortest distance between two points on Earth, a navigator needs to know two things: 1. The latitude of each point, which is its angular distance north or south of the equator.

P1

B Meridian

Now imagine a plane intersecting with the Earth perpendicular to the prime

H

North Pole O¢ b1 K

P1 and β1 represents its latitude. If α1 = 60.5° and β1 = 48.2°, we can say that point P1 has coordinates 60.5° N 48.2° W (i.e. 60.5° north of the equator and 48.2° west of the prime meridian). The distance (or arc measure) θ in radians between any two points north of the equator is given by the trigonometric formula θ = arccos(sin α1sin α2 + cos α1cos α2cos Φ) where

α1 = the longitude of point P1 α2 = the longitude of point P2, Φ = the change in latitude from P1 to P2, and

Although θ must be in radians, α1, α2 and Φ may be expressed in degrees.

P2 C

a1 O b1

Equator

A Prime meridian

South Pole

2. The longitude of each point, which is its angular distance east or west of the prime meridian. As an example, look at the figure above right. α1 represents the longitude of

Parallel

3 .3

EXERCISES

3. A triangle ABC has sides a, b and c such that

A. Trigonometric Theorems

a2 – b2 – c2 – (b ⋅ c) = 0. Find m(∠A) = α.

1. Find the length of side x in each triangle. a.

b.

A

4. A triangle ABC has sides a, b and c such that

A

30° 2

3

3 B

c2 – b2 = a2 + ñ2 ⋅ b ⋅ a. Find m(∠C) = α.

x

120°

C

4

B

5. Find the length x in each figure. a. A

x C

c.

d.

A

2

2

B

2

4

A

x

x

30°

A 8

B

2. Find the length x in each figure. b.

A

B

c.

3

e.

204

1 D

2

C

C

3

B

4

A

10

120°

30°

B

C

C

D

B

d.

B

E A

4 x

C

30°

a

9

a

E

3

a

A 6

2

ñ5

4

A

D 1 C

b.

A

a

D

1

c.

4

3

a. 6

x

f.

2

›

A 2

A x

6. Calculate sin α in each figure.

C

3

4

D

1

4

d.

x

B

B

B

A 3

D

x C

D A(DABC) = A(DDBC)

3

2

5

x

4

E

x

C

30°

A

4

2

C

C

6

1

x 60°

C

D

75°

B

c.

a.

A ñ3

x

C

60° B

45°

B

150°

4

b.

C

30° B

B

D 2BD = DC

C

ABCD is a square CE = BE

Algebra 10

7. The side lengths of a triangle ABC are 3, 7 and 8 units respectively. What is the circumference of the circumscribed circle of the triangle?

B. Trigonometric Formulas 13. Calculate the values without using a trigonometric table or a calculator.

8. In a triangle ABC, m(∠A) = 75°, m(∠B) = 60° and

a. sin 105°

b. cos 15°

c. tan 75°

d. cos 105°

e. tan 165°

f. cot 255°

g. sin 195°

h. cot 345°

AB = c = 10 cm. a. Find the length of the side b.

b. Find the radius of the circumscribed circle of the triangle.

9. In a triangle ABC,

2 1 and cos y = 3 4 value of each expression.

14. sin x =

A

m(∠ACB) = 120°, m(∠ABC) = 30° and AB = 6ñ3 cm. Find AC.

6ñ3 cm

30° B

are given. Find the

a. sin (x + y)

b. sin (x – y)

c. cos (x + y)

d. cos (x – y)

e. tan (x + y)

f. cot (x – y)

x 120° C

15. x and y are acute angles such that tan x = ›

3 . Evaluate the expressions. 5 b. cos (2x + y) a. tan (x + y)

1 and 4

tan y =

10. A triangle ABC has side lengths a = 3, b = 3ñ3 and c = 6. Its circumscribed circle has radius R = 3. Find the measure of each interior angle of the triangle.

c. sin (x + y)

d. cos (x – 2y)

e. cot (2x + 2y)

f. sin (x + 2y)

16. Simplify the expressions. ›

a. sin (x + 30°) + cos (x + 60°)

b. cos (x + y) + cos (x – y)

11. A triangle ABC has sides a = 7 cm, b = 24 cm and

c. sin (x + 30°) + sin (x – 30°)

c = 25 cm. Find A(ΔABC).

d. sin (x + y) – sin (x – y)

12. A triangle ABC has sides of length 8, 15 and 17 units respectively. Find the radius of its circumscribed circle. Trigonometry: Functions, Equations and Inequalities

17. Express cos 3α in terms of cos α and tan 3α in ›

terms of tan α. 205

18. Calculate sin 2x, cos 2x and tan 2x from the

21. Express each sum or difference as a product of

› information given in each question.

trigonometric functions.

3 π⎞ ⎛ and x ∈ ⎜ 0, ⎟ 5 2⎠ ⎝ 5 b. cos x = and csc x < 0 13 7 ⎛π ⎞ and x ∈ ⎜ , π ⎟ c. tan x = – 24 ⎝2 ⎠

a. sin x =

a. sin 5x + sin 3x

b. sin x – sin 4x

c. cos 4x – cos 6x

d. cos 9x + cos 2x

e. sin 2x – sin 7x

f. sin 3x + sin 4x

g. sin 11x + sin 9x

h. cos

5x x – cos 2 2

d. csc x = 4 and tan x < 0 e. cot x =

2 and sin x > 0 3

x x x and tan from the 2 2 2 › information given in each question.

19. Calculate sin , cos

a. sin x =

4 π⎞ ⎛ and x ∈ ⎜ 0, ⎟ 5 2⎠ ⎝

b. cos x = –

(Hint: 1 + cos x = cos 0° + cos x) x 2 x b. 1 – cos x = 2sin2 2 2 ⎛π c. 1+ sin x = 2 sin ⎜ + x ⎞⎟ ⎝4 2⎠

⎞ ⎠

c. csc x = 3 and x ∈ ⎜ , π ⎟

⎛π ⎝

3 ⎛ 3π ⎞ and x ∈ ⎜ , 2 π ⎟ 2 ⎝ 2 ⎠

⎛π ⎞ 2 sin ⎜ – x ⎟ 4 ⎝ ⎠ f. 1 – tan x = cos x

g. 20. Simplify each expression. a. sin 105° – sin 15°

b. cos 75° + cos 15°

c. cos 105° – cos 15°

d. sin 165° + sin 15°

e. sin 75° + sin 195°

f. sin 105° + sin 255°

g. cos π – cos 5π

h. sin 3π – sin π

206

12

⎠

⎛π ⎞ 2 sin ⎜ + x ⎟ 4 ⎝ ⎠ e. 1+ tan x = cos x

f. cot x = 5 and csc x < 0

12

x⎞

d. 1 – sin x = 2 cos 2 ⎜ + ⎟ 4 2

π⎞ ⎛ d. tan x = 1 and x ∈ ⎜ 0, ⎟ 2⎠ ⎝

e. sec x =

›

a. 1 + cos x = 2cos2

3 3π ⎞ ⎛ and x ∈ ⎜ π, ⎟ 5 2 ⎠ ⎝ ⎛π ⎝2

22. Verify each identity.

8

8

1 – tan x ⎛π ⎞ = tan ⎜ + x ⎟ 1+ tan x ⎝4 ⎠

⎛π ⎞ 2 sin ⎜ + x ⎟ 4 ⎝ ⎠ h. 1+ cot x = sin x ⎛π ⎞ 2 sin ⎜ – x ⎟ 4 ⎝ ⎠ i. 1+ cot x = sin x Algebra 10

23. Prove each statement.

26. In the figure, ABCD

›

cot α.

a. sin x + sin 5 x = tan 3 x cos x + cos 5 x

C

3 a

E 1

b. sin 3 x + sin 7 x = cot 2 x

A

cos 3 x – cos 7 x

c.

D

› is a rectangle. Find

2

F

B

4

sin 10 x cos 5 x = sin 9 x + sin x cos 4 x

d. sin x + sin 3 x+ sin 5 x = tan 3 x cos x + cos 3 x+ cos 5 x

27. In the figure, ABCD

e. sin x + sin y = tan x + y cos x + cos y

C

and BEFH are squares and AB = 2BE.

2

f.

sin ( x + y) – sin ( x – y) = tan y cos ( x + y)+ cos ( x – y)

g.

sin ( x + y) – sin ( x – y) = cot x cos ( x – y)+ cos ( x+ y)

24. In the figure, ABCD is

D

H

F

Find sin(∠AHE). A

A

B

E

2

C

D

a square with 3DE = 2EC and m(∠AEB) = θ. Find tan θ.

28. Find tan α in the E

q

A a

figure. 1 B

B

n

D

C

25. In the figure, CB ⊥ BA

y

and BC = 12. Find cos(∠OAC).

29. In the figure opposite,

C

A

ABC is an isosceles triangle. AB = AC, m(∠ABC) = θ , m(∠BAC) = α and

12

B(0, 4)

a

tan θ = 3. 0

x A(3, 0)

Trigonometry: Functions, Equations and Inequalities

Find tan α.

q B

C

207

30. In the figure, AD = 5 cm, BD = 13 cm, DF = 12 cm, DE = 4 cm and m(∠FDE) = α. Find cot α.

A 5 D

13

B

36. Evaluate

E a

1 1 + . sin 15 ° cos 15 °

4

12

F

C

37. Evaluate sin 10° ⋅ sin 30° ⋅ sin 50° ⋅ sin 70°.

››

31. sin 20° = x is given. Write sin 40° in terms of x.

38. cos 76° = x is given. Find sin 37° ⋅ cos 23° in terms of x. π 3 32. tan x – cot x = is given with 0 < x < . 2 4 › Find tan 2x – cot 2x.

39. x = 33. cos 2α = –0.125 is given with π < α < sin α .

π is given. Evaluate cos 8x ⋅ cos 2x. 12

3π . Find 2

40. π = 8α is given. Evaluate cos 5α ⋅ sin 3α. 34. Evaluate 3cos2 105° + sin2 15°.

π 4 is given and < α < π. 2 3 Evaluate (sin α ⋅ cos α) – cos2 α.

35. tan 2α =

208

π is given. Evaluate 6 › tan (2a – b) ⋅ tan (a – 2b).

41. a – b =

Algebra 10

In the previous section we looked at trigonometric identities. Recall that an identity in x is an expression which is true for all values of x. If an expression is only true for some values of x then it is called an equation. For example, sin2x + cos2x = 1 is true for all values of x, and so it is an identity. 1 is only true for x = 30°, x = 150° and their coterminal angles. So 2 1 sin x = is an equation. 2 In this section we will learn how to solve trigonometric equations.

However, sin x =

A. TYPES OF TRIGONOMETRIC EQUATION 1. Basic Equations a. sin x = a Remember! The y-axis is the sine axis.

Given the equation sin x = a, how can we find x? Look at the figure. For any number on the sine axis there are two corresponding values on the unit circle. One value is x1 = arcsin a and the other is x2 = π – arcsin a.

y a

p – arc sin a

arc sin a x

–x O

x

We should also consider the coterminal angles of x1 and x2, for example arcsin a + 2π, arcsin a + 4π, arcsin a + 6π, ... . We can write sin x = a ⇔ { x = arcsin a + 2kπ, k ∈ } or {x = π – arcsin a + 2πn, n ∈ } where –1 ≤ a ≤ 1. Alternatively, by studying this result carefully we can rewrite it in a shorter form: sin x = a ⇔ x = (–1)k arcsin a + kπ, k ∈ Trigonometry: Functions, Equations and Inequalities

for –1 ≤ a ≤ 1. 209

Special Results: When x = 0, x = 1 or x = –1 we can write the answer in a shorter way without using the formula. sin x = 0 ⇔ x = kπ, k ∈ . π sin x = 1 ⇔ x = + 2kπ, k ∈ . 2 π sin x = –1 ⇔ x = – + 2 kπ, k ∈ . 2

EXAMPLE

133 Solve sin x =

Solution 1

3 . 2

From the longer formula we have sin x =

3 3 3 + 2πn, n ∈ }. ⇔ {x1=arcsin +2 k π, k ∈ } and { x2 = π −arcsin 2 2 2

We know arcsin

π π 3 = 60 ° = . So x1 = + 2 k π, k ∈ . 2 3 3

π 2π Also x2 = ( π − ) + 2 nπ = + 2 nπ, n ∈ . 3 3

(1) (2)

2π π + 2 nπ}, k, n ∈ . Combining (1) and (2) gives us x = { + 2 kπ, 3 3

Solution 2

From the shorter formula we have sin x = a ⇔ x =( −1) k arcsin a + kπ, k ∈ . π So x = (–1) k + kπ, k ∈ . 3

Note Notice that both solutions give the same answer. However, the second, shorter form is more practical, since we can get the solutions by substituting only values of k.

EXAMPLE

134 Solve 4 sin 5x – 2 = 0.

Solution

210

2 1 = . 4 2 1 π π π kπ arcsin = 30 ° = , so the solution is 5 x= (–1) k + k π , i.e. x = (–1) k + , k∈ . 2 6 6 30 5

Rearranging the equation gives us sin5 x =

Algebra 10

EXAMPLE

135 Solve sin x = –

Solution

1. xn ⋅ xm = xn + m 2. –x = (–1) ⋅ x

arcsin(–

2 . 2

π 2 )= –45° = – , so we have 2 2

x =(–1)k·(–

π π π )+kπ = (–1)k·(–1)· +kπ, i.e. x = (–1)k+1 +kπ, k ∈ 4 4 4

b. cos x = a

y

If cos x = a, what is x?

Remember! The x-axis is the cosine axis.

x1 = arc cos a

Look at the figure. For any number on the cosine axis there are two corresponding values on the unit circle. One value is x1 = arccos a and the other is x2 = 2π – arccos a or x2 = – arccos a.

x O

–x

a

x

x2 = –arc cos a

We should also consider the coterminal angles of x1 and x2, for example arccos a + 2π, arccos a + 4π, .... . We can write cos x = a ⇔ x = ± arccos a + 2kπ, k ∈ where –1 ≤ a ≤ 1. Special Results: π cos x = 0 ⇔ x = + kπ, k ∈ . 2

cos x = 1 ⇔ x = 2kπ, k ∈ . cos x = –1 ⇔ x = π + 2kπ, k ∈ .

Note The points on the graph below show the solutions to the equation cos x = a. y

x

arc cos a – 2p

–arc cos a + 2p

–arc cos a – 2p –arc cos a

Trigonometry: Functions, Equations and Inequalities

arc cos a

–arc cos a + 4p arc cos a + 2p

211

EXAMPLE

1

136 Solve cos x = 2 .

Solution

We have cos x = a ⇔ x = ± arccos a + 2 k π, k ∈ . Also, arccos

1 π = 60 ° = . 2 3

1 π So x = ± arccos + 2 kπ = ± + 2 k π, k ∈ . 2 3

EXAMPLE

137 Solve 2cos x – ñ3 = 0.

Solution

Let us rewrite the equation in terms of cos x. Then we get cos x =

EXAMPLE

138 Solve cos x = –

Solution

EXAMPLE

3 3 π π , and we know arccos = 30 °= . So x = ± + 2 kπ, k ∈ . 2 2 6 6

arccos(–

2 3π 3π ) =135 ° = , so by the formula, x= ± + 2 k π, k ∈ . 2 4 4

139 Solve 2 cos 5x + 1 = 0.

Solution

Rewriting the equation gives us cos5 x= – So 5 x = ±

212

2 . 2

1 1 2π , and arccos( − ) =120 °= . 2 2 3

2π 2 π 2k π + 2 kπ, i.e. x = ± + , k∈ . 3 15 5 Algebra 10

EXAMPLE

140 Solve cos (3x) – 2 = 0.

Solution

EXAMPLE

Rewriting the equation gives us cos 3x = 2, but we know –1 ≤ cos 3x ≤ 1. So cos 3x = 2 is impossible. Therefore the equation has no solution.

π

141 Solve cos(3x – 4 ) = 0.

Solution

We know cos x = 0 ⇔ x = So 3 x –

EXAMPLE

π π + kπ, i.e. arccos 0 = . 2 2

3π π π π π π kπ = + kπ, which gives us 3 x= + + k π= + k π, i.e. x= + , k ∈ . 4 2 2 4 4 4 3

142 Solve sin( 2π – 3x) =1.

Solution

π We know that sin( – 3 x) = cos 3x, so we can solve cos 3x = 1 instead. 2 2kπ cos θ = 1 is a special result, so 3x = 2kπ and x = , k∈ . 3

c. tan x = a Remember! The red line in the figure is the tangent axis. y

x

If tan x = a, what is x?

y a

Look at the figure. For any number on the tangent axis there are two corresponding values on the unit circle. One value is x1 = arctan a and the other is the symmetry of x1 with respect to the origin, i.e. x2 = π + arctan a.

arc tan a

x O

x

(arc tan a) + p

By also considering the coterminal angles we can write tan x = a ⇔ x = arctan a + kπ, k ∈ for a ∈ . Trigonometry: Functions, Equations and Inequalities

213

EXAMPLE

143 Solve tan 3x = ñ3.

Solution

By the formula, 3 x = arctan 3 + kπ, k ∈ . We know π π π kπ arctan 3 = 60 ° = , so 3 x = + k π, i.e. x= + , k ∈ . 3 3 9 3

EXAMPLE

144 Solve 4tan(5x+ π3)+ 4 = 0.

Solution

π Rearranging the equation gives us tan(5 x + ) = –1. 3 π π π arctan(–1) = – 45 ° = – , so by the formula we have 5x + = – + kπ, i.e. 3 4 4 5x = –

EXAMPLE

7π 7 π kπ π π – + kπ = – + kπ and x = – + , k∈ . 4 3 12 60 5

145 Solve tan (2x – 13°) =

Solution Remember!

3 and write the answer in degrees. 3

3 = 30 °. Also, writing the solution formula in degree form gives us 3 2x – 13° = 30° + 180° k, i.e. 2x = 13° + 30° + 180° k = 43° + 180° k, so

We know arctan

43° +180 ° ⋅ k = 21.5 ° +90 ° ⋅ k, k ∈ . 2

We know that 1° = 60 minutes (60').

x=

So 0.5° = 30 minutes.

If we write 21.5° as 21°30' then the answer is

Instead of 21.5° we can write 21°30'.

x = 21°30' + 90° k, k ∈ .

d. cot x = a If cot x = a, what is x?

Remember! The red line in the figure is the cotangent axis.

By also considering the coterminal angles we can write

y

x

a arc cot a x O

x

(arc cot a) + p

cot x = a ⇔ x = arccot a + kπ, k ∈ for a ∈

214

y

Look at the figure. For any number on the cotangent axis there are two corresponding values on the unit circle. One value is x1 = arccot a and the other is the symmetry of x1 with respect to the origin, i.e. x2 = π + arccot a.

. Algebra 10

EXAMPLE

146 Solve 3 cot 2x = ñ3.

Solution

Rearranging the equation gives us cot 2 x = We know arccot

EXAMPLE

3 . 3

π kπ π 3 π = 60 ° = , so by the formula, 2x = + kπ and x = + , k ∈ . 3 6 2 3 3

147 Solve cot(– 2x) = –1.

Solution

x 3π 3π 3π + kπ and x = – – 2 k π, k ∈ . , so by the formula, – = 2 4 2 4 Note that since k is any integer it can be negative or positive, so we can also write arccot(–1) =135 ° =

x= –

3π 3π – 2 kπ = – + 2 k π, k ∈ . 2 2

Check Yourself 24 1. Solve the equations. a. (2 sin x) – 1 = 0

b. 7 sin x = 0

2 x π d. sin( – ) – =0 2 4 2

e. cos8 x = –

x c. (2 sin )+ 3 = 0 2 3 2

f.

π 1 cos( x + ) = – 6 2

2. Solve the equations. a. (2 tan 4x) + 2 = 0

π x b. tan( + ) = 3 4 2

x c. cot(– ) =1 2

π 3 b. sin( + 3 x) = 2 2

c. tan (π + 4x) = 0

d. cot 4x = ñ3 3. Solve the equations. 3π x 1 – )= 2 2 2 3π d. cot( – 2 x) =1 2

a. cos(

Trigonometry: Functions, Equations and Inequalities

215

4. Find the solutions of sin x = 5. Solve

1 in the interval 500º < x < 1000º . 2

sin5 x =1 . 3

Answers π 2π b. kπ, k ∈ c. (–1)k+1 + πk, k ∈ + 2 πk, k ∈ 6 3 π π 5π kπ 2 π d. (–1)k + + 2 πk, k ∈ e. ± f. ± π – + 2 πk, k ∈ + , k∈ 48 4 3 6 2 2

1. a. (–1)k

2. a. –

π π + n, n ∈ 16 4

b.

π 3. a. (–1)n+1 + 2 πn, n ∈ 3

π + 2 kπ, k ∈ 6

b. ±

π c. – + 2πk, k ∈ 2

2kπ π + , k∈ 18 3

c.

kπ , k∈ 4

d.

π kπ + , k∈ 24 4

d.

π kπ + , k∈ 8 2

4. 510°, 750°, 870° 5. no solution

2. Factorizing Equations We have seen how to solve simple trigonometric equations such as 7sin x = 0 and tan(4x + π) = 1. But how can we solve more complicated equations which contain a power of a ratio, or mixed ratios? One approach is to use factorization. If we can rewrite an equation in the factorized form a ⋅ b = 0 where a and b are trigonometric ratios, then we can solve a = 0 and b = 0 and find the union of the solution set. This union is the solution of the original equation. We can factorize an equation by using substitution and trigonometric identities and formulas.

EXAMPLE

148 Solve 2cos x – 3cosx + 1 = 0.

Solution

2

Let us factorize the expression. Let cos x = a, then the equation becomes 2a2 – 3a + 1 = 0, which we can factorize. 2a2 – 3a + 1 = (2a – 1) ⋅ (a – 1) = 0, which has solutions a = If a =

1 1 π then cos x = and so x = ± + 2kπ, k ∈ . 3 2 2

If a = 1 then cos x = 1 and so x = 2nπ, n ∈ .

(1) (2)

π The union of (1) and (2) gives us x ∈ {± + 2 kπ, 2 nπ}, k, n ∈ 3 equation. 216

1 and a = 1. 2

. This is the solution of the

Algebra 10

EXAMPLE

149 Solve sin x + sin x = 0. 2

Solution

Factorize the expression: sin x(sin x + 1) = 0 has solutions sin x = 0 and sin x = –1. If sin x = 0 then x = kπ, k ∈ . π If sin x = –1 then x = – + 2 nπ, n ∈ . 2 π Combining the two solutions gives us x ∈ {kπ, – + 2 nπ}, k, n ∈ . 2

EXAMPLE

150 Solve 3 tan x + 2 tan x = 1. 2

Solution

Making the right side zero gives us 3 tan2 x + 2 tan x – 1 = 0.

1 We can factorize this as (3 tan x – 1) ⋅ (tan x + 1) = 0, which has solution tan x = or 3 tan x = –1. 1 1 1 then x = arctan + kπ, k ∈ . ( arctan is not a special angle, so we will leave 3 3 3 it in this form.)

If tan x =

π If tan x = –1 then x = arctan(–1)+ nπ = – + nπ, n ∈ . 4 1 π In summary, the solution is x ∈ {arctan + kπ, – + n π}, k, n ∈ . 3 4

EXAMPLE

151 Solve sin 2x + sin x = 0.

Solution

Using the identity sin 2x = 2 sin x ⋅ cos x we can write 2 sin x ⋅ cos x + sin x = sin x (2cos x + 1) = 0, which means 1 sin x = 0 or 2cosx + 1 = 0 ⇒ cos x = – . 2

If sin x = 0 then x =kπ, k ∈ . 1 2π 2π 1 then arccos(– ) =120 ° = , so x= ± + 2 n π, n ∈ . 2 3 3 2 2π + 2 nπ}, k, n ∈ . In summary, the solution is the union: x ∈ {kπ, ± 3

If cos x = –

Trigonometry: Functions, Equations and Inequalities

217

EXAMPLE

152 Solve 2 sin x + 2 sin x = ñ2 + ñ2 sin x. 2

Solution

Let us factorize both sides separately: 2 sin2 x + 2 sin x = 2 sin x (sin x + 1), and ñ2 + ñ2 sin x = ñ2(1 + sin x). So we can rewrite the original equation as 2 sin x (sin x + 1) – ñ2(sin x + 1) = 0. Factorizing this gives (sin x + 1) ⋅ (2 sin x – ñ2) = 0, i.e. sin x = –1 or sin x =

2 . 2

π If sin x = –1 then x = – + 2 kπ, k ∈ . 2 2 π If sin x = then x = (–1)n + nπ, n ∈ . 4 2 π n π So the solution is x ∈ {– + 2 kπ, (–1) + n π}, k, n ∈ . 2 4 EXAMPLE

153 Solve cos 4x + cos 2x = 0.

Solution

We can factorize the left-hand side using the sum to product formulas: cos 4 x + cos 2 x = 2 ⋅ cos(

4x + 2 x 4x – 2 x ) ⋅ cos( ) = 2 cos 3 x ⋅cos x= 0. 2 2

So we have cos 3x = 0 or cos x = 0. π π kπ If cos 3x = 0 then 3x = + kπ and so x = + , k ∈ . 2 6 3 π If cos x = 0 then x = + nπ, n ∈ . 2 π kπ π So the solution is x ∈ { + , + n π}, k, n ∈ . 6 3 2

EXAMPLE

154 Solve

Solution

tan 3 x + tan 4 x 3 =– . 1 – tan 3 x ⋅ tan 4 x 3

tan x + tan y tan 3 x + tan 4 x , so is equivalent to 1 – tan x ⋅ tan y 1 – tan 3 x ⋅ tan 4 x 3 tan (3x + 4x). So we can rewrite the equation as tan7 x = – . 3

We know the sum formula tan( x + y) =

Since arctan(– 218

π π kπ 3 π ) = –30 ° = – , we have 7 x = – + kπ. i.e. x = − + , k ∈ . 3 6 6 42 7 Algebra 10

EXAMPLE

155 Solve sin (x + 20°) – sin (x – 20°) = 0.

Solution

By using the sum to product formulas we can write x + 20° + x – 20 ° x + 20 ° – x+ 20 ° ) ⋅ sin( ) 2 2 = 2 ⋅ cos x ⋅ sin 20 °. π So the equation is 2cos x sin 20° = 0 and so cos x = 0, i.e. x = + kπ, k ∈ . 2 sin( x + 20 °) – sin( x – 20 °) = 2 ⋅ cos(

EXAMPLE

156 Solve 2 cos x = 3 sin x. 2

Solution

One side of the equation contains the sine function and the other side contains the cosine function. We can change these to the same trigonometric ratio. Since cos2x = 1 – sin2x, we can rewrite the equation as 2 cos2x = 3 sin x, i.e. 2 (1 – sin2 x) = 3 sin x ⇒ 2 – 2 sin2x = 3 sin x ⇒ 2sin2x + 3 sin x – 2 = 0. Factorizing this gives us (2 sin x – 1 ) ⋅ (sin x + 2) = 0. So either π 1 ⇒ x = (–1) k + k π, k ∈ , or sin x = 2 which has no solution since –1≤ sin x ≤ 1. 2 6 π In conclusion, the solution is x = (–1)k + kπ, k ∈ . 6 sin x =

EXAMPLE

157 Find the sum of the roots of sin x+ cos

Solution

4

4

x=

7 in the interval [–π, π]. 8

Let us use the identity x2 + y2 = (x + y)2 – 2xy. Then we have 7 sin 4 x + cos 4 x = (sin 2 x+ cos 2 x) 2 – 2 sin 2 x ⋅cos 2 x= . 8

Using the identities sin2x + cos2x = 1 and (2 sin x ⋅ cos x)2 = sin2 2x we can write 12 –

7 1 sin 2 2 x sin 2 2 x 7 , i.e. = , which we can rearrange as 1 – = = 2 8 8 8 2

1 1 1 . So either sin 2 x= or sin 2 x= – . 4 2 2 π π kπ 1 If sin 2 x = then 2 x = (–1) k + kπ and so x= (–1) k + , k ∈ . 2 6 12 2 π π nπ 1 If sin 2 x = – then 2 x = (–1) n+1 + nπ and so x= (–1) n+1 + , n∈ . 2 6 12 2 π nπ π kπ + , (–1) n +1 + }, k, n ∈ . In summary, the solution is x ∈{( −1) k 12 2 12 2 sin 2 2 x =

Trigonometry: Functions, Equations and Inequalities

219

Practical Notation So far we have written the solution to a trigonometric equation as the union of other solution sets. However, this union might not be the shortest form of the answer. For example, one solution set may contain the other set. In this case, we only need to give the larger set as the answer. Alternatively, in a multiple-choice question we may find two or more answers but these may not be in the given choices. In this case it is helpful to draw all the answers on the unit circle. If the angles between the answers are the same then we will take the first non-negative answer and add the common angle multiplied by k ∈ . This will give us the answer.

y

y

y a

a

2p 3

p x

2p

p

one answer:

two answers:

x = a + 2πk

x = a + πk

x

2p 3

a x

2p 3

three answers: x = a+

2π k 3

y

y

a a

x

four answers: π x = a+ k 2

220

...

x

n answers: x = a+

2π k n

Algebra 10

EXAMPLE

158 Solve sin x ⋅ cos x = cos x.

Solution

Rearranging and then factorizing the equation gives cos x(sin x – 1) = 0, i.e. cos x = 0 or sin x = 1. π If cos x = 0 then x = + kπ, k ∈ Z. 2 π If sin x = 1 then x = + 2 nπ, n ∈ Z. 2 Inspect the answers:

π π 3π 5π + kπ = {± , ± , ± , ...}. 2 2 2 2 π π 3π 5π 7π 9π + 2 nπ = { , – , , – , , ...}. 2 2 2 2 2 2

We can see that the first set includes the second one, so we can write the final answer as π x = + kπ, k ∈ Z. 2

EXAMPLE

159 Solve 2 cos x – 1 = 0.

Solution

2

1 Rearranging the expression gives us cos 2 x = . 2

So either cos x =

If cos x =

2

=

2 1 2 or cos x = – =– . 2 2 2

2 π π 2 = 45 ° = then x = ± + 2 kπ, k ∈ . and arccos 4 2 4 2

If cos x = – then x = ±

1

2 3π 2 ) =135 ° = and arccos(– 2 4 2

–

p 4

y

p 4

3π + 2 nπ, n ∈ . 4

π 3π + 2 n π} where k, n ∈ . So x ∈ {± + 2 kπ , ± 4 4

Now sketch the answers on the unit circle, as

x 5p 4 –

p 4

shown at the right. We can see that a shorter π π form of the answer is x = + k, k ∈ Z. 4 2 Trigonometry: Functions, Equations and Inequalities

221

3. Equations with a Common Ratio Sometimes a trigonometric equation includes the same ratio on each side of the equation, for example cos x = cos 30°. We can use the following solution formulas in this case. 1. If cos x = cos a then x = a + 2kπ or x = –a + 2kπ, k ∈ . 2. If sin x = sin a then x = a + 2kπ or x = (π – a) +2kπ, k ∈

⇒ x = (–1)ka + kπ, k ∈ .

3. If tan x = tan a then x = a + kπ, k ∈ . 4. If cot x = cot a then x = a + kπ, k ∈ . Note that we may need to use trigonometric identities, factorization or substitution to bring an equation to one of these simple forms.

EXAMPLE

160 Solve cos x = cos 30°.

Solution 1

From the formula above, x = 30° + 360°k or x = –30° + 360°k, k ∈ . π We can also write the answer in radians as x = ± + 2kπ , k ∈ . 6

Solution 2

Let us try to solve the equation without using the formula. We can write cos x – cos 30° = 0. Using the sum to product formulas gives us x + 30° x – 30 ° ) ⋅sin( ) = 0, i.e. 2 2 x + 30° x – 30 ° sin( ) = 0 or sin( ) = 0. 2 2 x + 30° x + 30 ° If sin( ) = 0 then = kπ, so x= –30 °+ 360 °k, k ∈ . 2 2 x – 30 ° x – 30 ° If sin( ) = 0 then = nπ, so x= 30 °+ 360 °n, n ∈ . 2 2 So the final ans wer is x ∈ {–30 °+ 360 °k, 30° + 360 °n}, k, n ∈ . cos x – cos 30 ° = −2 ⋅ sin(

EXAMPLE

161 Solve sin 2x = sin 20°.

Solution 1

By the solution formula, 2x = 20° + 360°k or 2x = (180° – 20°) + 360°n. So x = {10° + 180°k, 80° + 180°k}, k ∈ .

Solution 2

By the second formula for sin x = sin a, 2x = (–1)k ⋅ 20° + 180°k and so x = (–1)k ⋅ 10° + 90°k, k ∈ .

222

Algebra 10

EXAMPLE

cos x

cos 2 x

162 Solve cos5x = cos6 x .

Solution

Cross multiplying gives us cos x ⋅ cos 6x = cos 5x ⋅ cos 2x. By the product to sum formulas on each side of the equation we get 1 1 [cos( x +6 x)+ cos( x – 6 x) ]= [cos(5 x+ 2 x)+ cos(5 x – 2 x) ]. This simplifies to 2 2

cos 7x + cos (–5x) = cos 7x + cos 3x. Rearranging the terms gives us cos( − x) = cos x

cos 5x – cos 3x = 0, i.e. –2 sin(

5 x + 3x 5x – 3x )sin( )= 0 2 2

⇒ –2 ⋅ sin 4x ⋅ sin x = 0. So

sin 4x = 0 or sin x = 0. If sin 4x = 0 then 4x = kπ and so x =

kπ , k∈Z . 4

If sin x = 0 then x = nπ, n ∈ . We know that {nπ , n ∈ } ∈ { x∈{

EXAMPLE

π

π

163 Solve tan(4x+ 3 ) = cot(2 x – 5).

Solution

cot x = tan (

kπ }, k ∈ Z. 4

kπ , k ∈ } . So the final solution is 4

π 2

One side contains a tangent but the other side is a cotangent. We need to rewrite one ratio in terms of the other to get a simpler equation. π π π 7π On the right-hand side we can write cot(2 x – ) = tan( – 2 x+ ) = tan( – 2 x). 5 2 5 10

– x)

π 7π So the equation becomes tan(4 x + ) = tan( – 2 x). By the solution formula for tan x, 3 10 4x +

π 7π – 2 x + k π. Rearranging the terms gives us = 3 10

6x =

7π π 11π 11π kπ – + kπ = + k π, i.e. x ∈{ + }, k ∈ Z. 10 3 30 180 6

Trigonometry: Functions, Equations and Inequalities

223

EXAMPLE

164 Solve sin 5x + sin 4x = sin 3x + sin 2x.

Solution

2

2

2

2

We cannot factorize the equation directly, but we can rewrite it as sin2 5x – sin2 2x = sin2 3x – sin2 4x. Now we can factorize: (sin 5x – sin 2x)(sin 5x + sin 2x) = (sin 3x – sin 4x)(sin 3x + sin 4x). By the sum to product formulas on both sides we get 2 cos(

5x + 2 x 5x – 2 x 5x + 2 x 5x – 2 x ) ⋅ sin( ) ⋅ 2 sin( ) ⋅cos( )= 2 2 2 2 3x + 4 x 3x – 4 x 3x + 4 x 3x – 4 x 2 cos( ) ⋅ sin( ) ⋅ 2 sin( ) ⋅cos( ). 2 2 2 2

This simplifies to 2 cos

7x 3x 7x 3x 7x 7x x x ⋅ sin ⋅ 2 sin ⋅ cos ⋅sin( − ) ⋅2 sin ⋅cos( − ). = 2 cos 2 2 2 2 2 2 2 2

By using the identity 2 sin x ⋅ cos x = sin 2x we can simplify this further to sin 7x ⋅ sin 3x = sin 7x ⋅ sin (–x) = –sin 7x ⋅ sin x. So sin 7x ⋅ sin 3x + sin 7x ⋅ sin x = 0, i.e. sin 7x (sin 3x + sin x) = 0 ⇒ sin 7x = 0 or sin 3x + sin x = 0. If sin 7x = 0 then 7x = kπ ⇒ x =

kπ , k ∈ Z. 7

(1)

If sin 3x + sin x = 0 then sin 3x = –sin x = sin (–x). By the solution formula for sine we have nπ a. 3x = –x + 2nπ, i.e. 4x = 2nπ ⇒ x = , n ∈ Z. 2

(2)

b. 3x = (π – (–x)) + 2mπ = π + x + 2mπ, i.e. 2x = π + 2mπ ⇒ x =

π + mπ , m ∈ Z. (3) 2

As we can easily see, (3) is a subset of (2), so (2) is a sufficient answer. Finally, combining kπ nπ }, k, n ∈ Z. (1) and (2) gives us x ∈ { , 7 2

Check Yourself 25 1. Solve the equations.

224

a. cos2x + cos x = 2

b. 4 cos2x – 3 cos x = 0

d. sin 2x = sin x

e.

2 ⋅ tan 3 x 3 = 3 1 – tan 2 3 x

c. 3 cos 2x – 2 cosx + 3 = 0 f. cot2x – 3 cot x + 2 = 0 Algebra 10

2. Write the solution to each equation in its shortest form. a. tan2x – 1 = 0

b. 2cos2x + cos x – 1 = 0

x 5x 5x 3. Find the sum of the roots of 2 sin sin – sin = 0 in the interval [0, 2π]. 2 2 2

4. Solve each equation. a. sin 5x = sin 3x

b. cos 3x – cos x = 0

d. cos 3x = sin x

π 2π e. sin(2 x + ) = sin( x – ) 3 3

c. tan 2x + tan 3x = 0

Answers π 3 b. { + kπ, ± arccos + 2 nπ}, k, n ∈ Z 2 4

1. a. 2kπ, k ∈ π 2 c. { + kπ, ± arccos + 2 n π}, k, n ∈ Z 2 3

e. 2. a.

kπ π + , k∈Z 36 6 π kπ + , k∈Z 4 2

π d. {kπ, ± + 2 nπ}, k, n ∈ Z 3 π f. { + kπ, arccot2+ nπ}, k, n ∈ Z 4

b.

π 2 kπ + , k∈Z 3 3

3. 8π 4. a. {kπ,

π nπ + }, k, n ∈ Z 8 4

π kπ π d. { + ,– + nπ}, k, n ∈ Z 8 2 4

b. {kπ,

nπ }, k, n ∈ Z 2

e. {π + 2 kπ,

c.

kπ , k∈Z 5

4π 2 nπ + }, k, n ∈ Z 9 3

4. Linear Equations in sin x and cos x Recall that ax + by = c is a linear equation in the two variables x and y. If we replace x and y with sin x and cos x then we obtain a linear equation in sin x and cos x. In other words, the equation a sin x + b cos x = c is a linear equation in sin x and cos x. a b or cot α = . Then b a we can use the sum and difference formulas to simplify the given expression. Finally, we can

To solve a linear equation like this, we can use the substitution tan α =

use the methods we have already seen to solve the question. Trigonometry: Functions, Equations and Inequalities

225

First let us try to get

b from the given equation. For this we will divide both sides of the equaa

tion by a: a cos x + b sin x = c ⇒

a cos x + b sin x c = , i.e. a a

b c cos x + sin x = a a cos x +

sin α c sin x = cos α a

(use tan α =

cos x ⋅ cos α + sin α ⋅ sin x c = cos α a

sin α b b and we know that tan α = = ) a cos α a

(equalize the denominators)

cos( x – α ) c c = , i.e. cos( x – α) = cos α. a cos α a

(by the difference formula for cosine) A

Now draw a right triangle and let 2

a b . So tan α = , so cos α = 2 a a + b2 cos( x – α) =

c c a cos α = ⋅ = 2 a a a + b2

2

a

c 2

a + b2

.

+b

b

a B

a

C

In conclusion, to solve a linear equation of the form asin x + bcos x = c we can compare the values of c2 and a2 + b2. 1. If c2 > a2 + b2 then cos( x − α) = equation.

c 2

a + b2

> 1 and so there is no solution for the linear

2. If c2 = a2 + b2 then cos (x – α) = 1 and so x = α + 2kπ, k ∈ . 3. If c2 < a2 + b2 then we can solve the equation by using the methods that we have learned for equations with a common ratio. EXAMPLE

165 Solve 2 sin x + 3 cos x = 4.

Solution

We have a sin x + b cos x = c with a = 3, b = 2 and c = 4. Let us compare c2 and a2 + b2. We get 22 + 32 = 4 + 9 = 13 < 42 = 16. Since c2 > a2+ b2, by the above result we can say that the equation has no solution.

EXAMPLE

166 Solve cos x + ñ3 sin x = 2.

Solution

We have a = 1, b = ñ3 and c = 2. Let us compare c2 and a2 + b2. We get a2 + b2 = 12 + ñ32 = 1 + 3 = 4 = c2. Then by the above result we can find the solution. First find α : tan α =

226

b 3 π π π = = tan , i.e. α = so the solution is x = + 2kπ, k ∈ . a 1 3 3 3 Algebra 10

EXAMPLE

167 Solve ñ3 cos x + sin x = 1.

Solution 1

We have a = ñ3, b = 1 and c = 1. a2 + b2 = ñ32 + 12= 3 + 1 = 4 > 1 = c2, so we will use the equations with a common ratio. tan α =

b 1 3 π π = = = tan ⇒ α = . So by the above result we have a 3 6 6 3

cos( x – α) =

π 1 π 1 3 1 c cos = ⋅ = . cos α and cos( x − ) = 6 6 2 2 a 3 3

π 1 So the equation becomes cos( x – ) = . 6 2 1 π π π π π Since arccos = 60 ° = , we have x – = ± + 2 kπ and so x = ± + 2 kπ, k ∈ Z. 6 3 6 3 2 3

Solution 2

a b we can use cot α = . a b The equation is ñ3cosx + sinx = 1, and dividing both sides by a = ñ3 gives us

In this type of question, instead of using the substitution tan α =

3 3 3 cos α 3 , i.e. = . sin x = . Now let us use the substitution cot α = 3 3 3 sin α 3

cos x +

By using the sum formula for sin(x + α) we get sin( x + α ) = cot α = arcsin

EXAMPLE

3 ⋅ sin α. 3

3 π π 3 3 1 , i.e. α = . So sin( x + ) = ⋅ = . 3 3 3 3 2 2

1 π π π π π = , so the answer is x + = ( −1)k + kπ or x = ( −1) k − + kπ, k ∈ . 2 6 3 6 6 3

168 Solve 3 sin x + ñ3 cos x = 3.

Solution

We have a=ñ3 , b= 3 and c = 3. We get a2 + b2 = ñ32 + 32= 3 + 9 = 12 > 9 = c2, so we will use the equations with a common ratio. tan α =

π π 3 b = = 3 = tan so α = . 3 3 a 3

π π 3 3 1 3 ⋅ = cos( x − ) = cos = . 3 3 2 3 3 2 Since arccos

π π π π 3 π = , we have x − = ± +2 kπ ⇒ x = ± +2 k π, k ∈ . 2 6 3 6 3 6

Trigonometry: Functions, Equations and Inequalities

227

EXAMPLE

169 Solve cos x + sin x = 1.

Solution

We have a= 1 , b= 1 and c = 1. We get a2 + b2 = 12 + 12= 2 > 1 = c2, so a2 + b2 > c2 and we will use equations with a common ratio. π π b 1 ⇒ α = . tanα= = = 1 = tan 4 4 a 1 π π 1 2 c cos( x − α) = cos α, i.e. cos( x − ) = ⋅cos = . 4 1 4 2 a π π π π 2 π = , we have x − = ± + 2 kπ, i.e. x = ± +2 k π, k ∈ . Since arccos 2 4 4 4 4 4 We can also write these answers separately: π x1 = 2kπ, k ∈ and x2 = + 2 n π, n ∈ . 2

5. Homogeneous Equations in sin x and cos x Recall that the degree in a term in an equation is its power: 2x3 has power 3. Definition

homogeneous equation If the degrees of all the terms in an equation are the same, the equation is called a homogeneous equation. For example, ax + by = 0 is a first-order homogeneous equation, ax2 + bxy + cy2 = 0 is a second-order homogeneous equation, and so on. If we substitute sin x and cos x for x and y in a homogeneous equation, the equation becomes a homogeneous equation in sin x and cos x. In this section we will look at the solutions of homogeneous equations in sin x and cos x.

a. First-order homogeneous equations Let a cos x + b sin x = 0 be a given equation. Dividing both sides by cos x gives us a cos x + b sin x 0 = (cos x ≠ 0). This simplifies to cos x cos x a a sin x ⇒ x = arctan(– ) + k π, k ∈ . a+ b = 0 , i.e. a + b tan x = 0 ⇒ tan x = – b b cos x 228

Algebra 10

EXAMPLE

170 Solve sin x + cos x = 0.

Solution

Dividing both sides by cos x (cos x ≠ 0) gives us

sin x +1= 0 . cos x

We can rewrite this as tan x = –1. Since arctan(–1) = –45 ° =

EXAMPLE

π π , we have x = + kπ, k ∈ . 4 4

171 Solve ñ3 cos x – 3 sin x = 0.

Solution

Let us divide both sides by cos x (cos x ≠ 0). Then we have 3–3

sin x 3 = 0, i.e 3 – 3tan x = 0 ⇒ tan x= . cos x 3

Since arctan

3 π π = 30 ° = , the solution is x= + kπ, k ∈ Z. 3 6 6

b. Second-order homogeneous equations Let a cos2 x + (b cos x ⋅ sin x) + c sin2 x = 0 be a given equation. Dividing both sides by cos2 x gives us 0 a cos 2 x + b cos x ⋅ sin x + c sin 2 x = (cos x ≠ 0), which simplifies to 2 cos x cos 2 x sin x sin 2 x +c = 0, i.e. a + b tan x+ c tan 2 x = 0. a+ b cos x cos 2 x

This is a quadratic equation in tan x. We can now find the solution using factorization. Trigonometry: Functions, Equations and Inequalities

229

EXAMPLE

172 Solve cos x – (3 cos x ⋅ sin x) – 4 sin x = 0.

Solution

2

2

Let us divide both sides by cos2 x (cos x ≠ 0). Then we have cos 2 x – 3cos x ⋅ sin x – 4sin 2 x 0 sin x sin 2 x = , i.e 1 – 3 – 4 = 0. We can rewrite this as cos x cos 2 x cos 2 x cos 2 x 1 – 3 tan x – 4 tan2 x = 0.

Factorizing this equation gives us (1 – 4 tan x) ⋅ (1 + tan x) = 0. There are two cases: 1 1 a. If 1 – 4 tan x = 0 then tan x = and so x = arctan + kπ, k ∈ Z. (1) 4 4 π b. If 1 + tan x = 0 then tan x = –1 and so x = – + nπ, n ∈ Z . (2) 4 1 π Combining (1) and (2) gives us the final answer: x = {arctan + kπ, – + nπ}, k, n ∈ Z. 4 4

6. Maximum and Minimum Values of a Trigonometric Function A Let f(x) = a cos x + b sin x for a, b ∈ . How can we find the maximum and b minimum values of f(x)? a+ b sin α b Let tan α = = . a cos α Let us draw a right triangle with an acute a angle α. We can write a B C f(x) = a cos x + b sin x (take α out of the paranthesis) b = a(cos x + sin x) sin α a (use the substitution tanα = ) cos α sin α = a(cos x + sin x) cos α cos x ⋅ cos α + sin x ⋅ sin α (eqalize the denominators and use the sum formulas) a = a( ) (from the right triangle we have cosα = ) cos α 2 a + b2 a cos( x – α) = a 2

2

a2 + b 2 = a2 + b2 cos( x – α).

We know –1 ≤ cos(x – α) ≤ 1. Multiplying both sides by

a2 + b2 gives us

– a2 + b2 ≤ a2 + b 2 cos( x – α) ≤ a 2 + b 2 . In summary,

maximum value of f ( x) = a2 + b 2 . minimum value of f ( x) = – a2 + b 2 . 230

Algebra 10

EXAMPLE

173 Find the maximum and minimum values of f(x) = 3 cos x – 4 sin x.

Solution

We are given a = 3 and b = –4. By the formula we have just seen, maximum value of f ( x) = a2 + b 2 = 32 +(–4)2 = 9+16 = 25 = 5 minimum value of f ( x) = – a2 + b2 = – 32 +(–4)2 = – 9+16 = – 25 = –5.

EXAMPLE

174 Find the maximum and minimum values of f(x) = sin x + cos x.

Solution

We have a = 1 and b = 1. By the previous formula: maximum value of f ( x) = a2 + b 2 = 12 +12 = 1+1 = 2 minimum value of f ( x) = – a2 + b 2 = – 12 +12 = – 1+1 = – 2.

EXAMPLE

175 Find the range of the function f(x) = sin x + ñ3 cos x.

Solution

We know that sin x and cos x are continuous functions, so the range will be an interval between the maximum and minimum values of f(x). By using the previous formula with a = ñ3 and b = 1 we have 2 2 maximum value of f ( x) = a + b = 2 2 minimum value of f ( x) = – a + b = –

2

3 +12 = 3+1 = 4 = 2 2

3 +12 = – 3+1 = – 4 = –2.

So the range is [–2, 2]. Trigonometry: Functions, Equations and Inequalities

231

EXAMPLE

176 Find the value of x that gives A = (sin x + 3) ⋅ (1 – sin x) its maximum value.

Solution 1

First we will find the maximum value of A, then we will find the value of x at this point. A = (sin x + 3) ⋅ (1 – sin x) = 3 – 2 sin x – sin2 x = 4 – (1 + 2 sin x + sin2 x) = 4 – (1 + sin x)2. We know (1 + sin x)2 ≥ 0, so A = 4 – (1 + sin x)2 ≤ 4. So the maximum value of A is 4. Now we need to solve the equation (sin x + 3) ⋅ (1 – sin x) = 4 for x: 3 – 2 sin x – sin2 x = 4 ⇒ sin2 x + 2 sin x + 1 = 0. This factorizes to (sin x + 1)2 = 0, i.e. sin x = –1. So the answer is π x = – + 2 kπ , k ∈ Z. 2

Solution 2

We can get the maximum value of a function f(x) = a ⋅ b when a = b. So A = (sin x + 3) ⋅ (1 – sin x) is at its maximum when sin x + 3 = 1 – sin x. This gives us 2 sin x = 1 – 3 = – 2, i.e. sin x = –1. So the answer is π x = – + 2 kπ , k ∈ Z. 2

Check Yourself 26 1. Solve the equations. a. 4 sin x + 3 cos x = 6

b. ñ3 sin x + cos x = 1

c. 2sin x + ñ5 cos x = 3

b. sin x – ñ3 cos x = –1

c. sin x – cos x = 1

2. Solve the equations. a. 3cos x + ñ3 sin x = 3

3. Solve the first-order homogeneous equations. a. sin x + ñ3 cos x = 0

b. sin x + 2 cos x = 0

c. 3 sin x – cos x =0

4. Solve ñ3 cos2 x – (1 + ñ3)sin x ⋅ cos x + sin2 x = 0. 5. Find the maximum and minimum values of each trigonometric function. a. 2 sin x + 3 cos x

b. 5 sin x – 12 cos x

c. 3 cos x – 4 sin x

6. Find the range of f(x) = 4 sin x + 3 cos x + 2. 232

Algebra 10

Answers 1. a. no solution

b. {

2π + 2 kπ,2 nπ}, k, n ∈ Z 3

c. arctan

2 5 + 2 kπ, k ∈ Z 5

π 2. a. { + 2 kπ, 2 nπ}, k, n ∈ Z 3 π π b. {– + 2 kπ, + 2 nπ}, k, n ∈ Z 2 6 π c. { + 2 kπ, π + 2 nπ}, k, n ∈ Z 2 π 3. a. – + kπ, k ∈ Z 3

1 c. (arctan )+ kπ, k ∈ Z 3

b. (– arctan 2)+ kπ, k ∈ Z

π π 4. { + kπ, + nπ}, k, n ∈ Z 4 3

5. a. ò13, – ò13

b . 13, –13

c . 5, –5

6. [–3, 7]

B. FURTHER TRIGONOMETRIC EQUATIONS 1. Equations with an Absolute Value If a trigonometric equation includes an absolute value, we can solve it by considering the possible values of the absolute expression. EXAMPLE

177 Solve |3 cos 5x + 2| – 3 = 0.

Solution

Rearranging the equation gives us |3 cos 5x + 2| = 3. So either 3 cos 5x + 2 = 3 or 3 cos 5x + 2 = –3. 1. If 3 cos 5x + 2 = 3 then 3 cos 5x = 1, i.e. cos5 x = So x = ±

1 1 2 kπ arccos + , k ∈ Z. 5 3 5

1 1 and 5x = ± arccos + 2 kπ . 3 3

5 2. If 3 cos 5x + 2 = –3 then 3 cos 5x = –5, i.e. cos 5x = – < –1 . This equation has no 3 solution.

In conclusion, x = {± Trigonometry: Functions, Equations and Inequalities

1 1 2 kπ arccos + }, k ∈ Z is theonly solution. 5 3 5 233

EXAMPLE

178 Solve |sin x + cos x| = sin x – cos x.

Solution

Consider the two possibilities for the absolute value: 1. If sin x + cos x = sin x – cos x then 2 cos x = 0, i.e. cos x = 0. So π x = + 2kπ, k ∈ Z. 2 2. If sin x + cos x = –(sin x – cos x) = –sin x + cos x then 2 sin x = 0, i.e. sin x = 0. So x = nπ, n ∈ .

π So the solution is x ∈ { + 2kπ, nπ}, k, n ∈ Z. 2

2. Parametric Equations If an equation includes one or more variables which may affect the result of the equation, the equation is called a parametric equation. EXAMPLE

179 How many solutions does the equation sin x + sin x = m have in [0, 2π], depending on m?

Solution

2

Let us begin by finding the maximum and minimum values of sin2 x + sin x. If we substitute t = sin x then the expression is equivalent to the function y = t2 + t. 1 b The vertex of y is V(r, k) where r = – = – and 2a 2

y = ax2 + bx + c is the form of a parabola. The vertex of a parabola is V(r, k), where b and r=– 2a k =y(k).

234

1 1 1 1 1 1 1 k = y(– ) = (– )2 + (– ) = – = – . So – is the minimum value of y. 2 2 2 4 2 4 4

To find the maximum value of y we will substitute the maximum value of sin x into y. The maximum value of sin x is 1, so the maximum value of y is 12 + 1 = 2. 1 1 This means – ≤ y ≤ 2, and when y < – and y > 2 there is no solution. 4 4 We can rewrite the equation t2 + t = m as t2 + t – m = 0. Let us find the discriminant of this quadratic equation: Δ = b2 – 4ac = 12 – 4 ⋅ 1 ⋅ (–m) = 1 + 4m. 1 1. If Δ < 0, i.e. m < – , there is no real solution. 4 1 2. If Δ = 0, i.e. m = – , there is one root for the quadratic equation: 4 1 1 1 1 t2 + t + = 0 = ( t + )2 = 0 and t = – . So sin x= – and x ∈[0, 2 π]. 4 2 2 2 1 So for sin x = – we have two answers: x = 210° and x = 330°. 2 1 3. If Δ>0, i.e. 2 ≥ m > – there are two roots, and for every root there are two values of sin x. 4 1 In summary, the equation has no solution when m < – and m > 2. It has two solutions if 4 1 1 m = – , and it has a pair of solutions for each value of m in ( − ; 2] . 4 4 Algebra 10

3. Equations with Inverse Trigonometric Functions If an equation includes inverse trigonometric functions such as arcsin, arccos, arctan, etc., we can solve it by using the rules of inverse trigonometric functions, i.e. if arcsin x = y then sin y = x, if arctan x = y then tan y = x, etc. EXAMPLE

4

180 Solve arctan x+ arctan(1 – x) = arctan 3.

Solution

In this type of problem, first we try to convert inverse trigonometric functions to regular trigonometric functions so that the equation is easier to solve. We can work on each term separately: If arctan x = α then tan α = x. If arctan (1 – x) = β then tan β = 1 – x. 4 4 If arctan = y then tan y = . 3 3 4 Now we can write arctan x + arctan (1 – x) = arctan as α + β = y. 3 Let us take the tangent of both sides, then tan (α + β) = tan y. We can rewrite this as tan α + tan β x +1 – x 1 4 = tan y, i.e. = = , i.e. 2 1 – tan α ⋅ tan β 1 – x ⋅(1 – x) 1 – x + x 3 3 = 4 – 4 x + 4 x2 or 4 x2 – 4 x+1= 0. This is the square of (2x – 1). So 1 (2 x – 1)2 = 0 ⇒ 2 x – 1= 0, i.e. x = . This is the solution of the equation. 2

4. Systems of Trigonometric Equations A system of trigonometric equations includes two or more trigonometric equations. To solve a system of equations we either eliminate one of the trigonometric ratios by adding or subtracting, or we find one ratio in terms of the other and then use substitution. EXAMPLE

x sin a + 3sin b = cos a

181 Solve the system of equations ⎧⎨⎩x cos a + 3cos b = – sin a .

Solution

We want to eliminate one of the trigonometric ratios. For this the coefficients must be the same. So let us multiply the first equation by cos b and the second equation by –sin b, respectively. Then x sin a cos b + 3sin b cos b = cos a cos b – x cos a sin b – 3 sin b cos b = sin a sin b x(sin a cos b – cos a sin b) = cos acos b+ sin asin b x ⋅ sin( a – b) = cos( a – b) x=

Trigonometry: Functions, Equations and Inequalities

(add the two equations) (by the sum and difference formulas)

cos( a – b) = cot( a – b ). sin( a – b) 235

EXAMPLE

1

182 Solve the system of equations ⎧⎪⎪⎨sin x cos y = 2

Solution

. ⎪cos x sin y = – 1 2 us Adding and subtracting the two⎪⎩ expressions gives 1 1 sin x cos y + cos x sin y = – = 0 , i.e. sin (x + y) = 0 ⇒ x + y = kπ, and 2 2 1 1 π sin x cos y – cos x sin y = + =1 , i.e. sin (x – y) = 1 ⇒ x – y = + 2 nπ. 2 2 2

(1) ⎧ x + y = kπ Now we have another system of equations: ⎪⎨ π (2) ⎪⎩ x – y = 2 + 2nπ. kπ π π Adding these equations gives us 2 x = + 2 nπ + kπ, i.e. x = + nπ + , k, n ∈ Z. 2 4 2 π π kπ kπ = − − nπ + , k, n ∈ Z. − nπ − 4 2 4 2 π kπ π kπ In summary, the solution is ( x, y) = ( + nπ + , − − nπ + ), k, n ∈ Z. 4 4 2 2 From (1) we can write y = kπ – x = kπ −

EXAMPLE

183 Solve the system of equations ⎧⎪⎨sin x+ sinπ y = –

Solution

2

. ⎪x + y = 2 ⎩ π From the second equation we get y = – x . Let us use this in the first equation: 2 π sin x + sin( – x) = – 2 , i.e. 2

sin x + cos x = –ñ2 ⇒ cos x + 1 ⋅ sin x = –ñ2.

π Substitute 1= tan = 4

(1)

π 4 then (1) becomes π cos 4 sin

π π π cos x cos + sin xsin 4 sin x = 4 4 = – 2, i.e. cos( x – π ) = – 2 ⋅cos π = – 2 ⋅ 2 = –1. cos x + π π 4 4 2 cos cos 4 4 We know arccos( −1) = π, s o 5π π π π + 2 kπ, k ∈ Z. x – = x − = π + 2kπ and x = + π + 2 kπ = 4 4 4 4 3π π π 5π – 2 kπ = – – 2 kπ, k ∈ Z. y= – x= – 2 2 4 4 3π 5π – 2 kπ ), k ∈ Z. In summary, ( x, y) = ( + 2 kπ ,– 4 4 sin

236

Algebra 10

5. Mixed Examples We have now seen several different techniques which we can use to solve trigonometric equations: 1. We can factorize the terms.

sin α . cos α 3. We can use trigonometric identities and formulas.

2. We can use a substitution such as tan α =

4. We can consider different parts of an equation separately, and combine individual results to get the final solution. There are no strong rules about which technique is appropriate for which problem. Often we may try different approaches before we find the answer. The best way to improve your problem-solving skills is to practise on many different types of problem. In this section we will look at some mixed problems and their solutions. EXAMPLE

184 Solve cos 2x = cos x – sin x.

Solution

The equation contains both sin x and cos x, so it is not clear what we should write instead of cos 2x. Let us try using the identity cos 2x = cos2 x – sin2 x. Then the equation becomes cos2 x – sin2 x = cos x – sin x. Factorizing the left side gives (cos x – sin x) ⋅ (cos x + sin x) – (cos x – sin x) = 0, i.e. (cos x – sin x) ⋅ (cos x + sin x – 1) = 0. There are two cases: cos x 1. If cos x – sin x = 0 then cos x = sin x and so 1= = tan x. sin x π So x = + kπ , k ∈ Z. π 4 sin π 4 2. If cos x + sin x = 1 then we can use the substitution 1= tan = π 4 cos 4 π sin 4 sin x =1, i.e. cos x ⋅cos π+ sin x ⋅ sin π = cos π= 2. to get cos x + π 4 4 4 2 cos 4 π 2 This simplifies to cos( x – ) = . 4 2 π π π π From the solution formula for cosine, x – = ± + 2 nπ and so x = ± + 2 n π, n ∈ Z. 4 4 4 4 π π π In summary, x ∈ { + kπ , ± + 2 n π}, k, n ∈ Z. 4 4 4

EXAMPLE

185 Find the sum of all the roots of sin1 3x + cos1 3x = 163 in [0, 2π ].

Trigonometry: Functions, Equations and Inequalities

2

2

237

Solution

Equalize the denominators on the left-hand side:

1 1 cos 2 3 x +sin 2 3 x + = sin 2 3 x cos 2 3 x sin 2 3 x ⋅ cos 2 3 x

1 4 4 = = . So the equation becomes 2 2 2 sin 3x ⋅ cos 3 x 4sin 3 x ⋅cos 3 x sin 2 6 x 4 16 3 3 3 = , i.e. sin 2 6 x = and so sin6 x = or sin6 x =– . 2 4 2 2 sin 6 x 3 π π kπ 3 1. If sin6 x = + , k∈ . then 6 x = (–1) k + kπ and so x =(–1) k 2 3 18 6 π π 7π 4π k = 0 gives x = , k = 1 gives x = , k = 2 gives x = , k = 3 gives x = . 18 9 18 9

=

2

3 k π nk then 6 x = (–1) k+1 + nπ and so x= ( −1) k+1 + , n ∈ Z. 2 3 18 6 5π 2π k = 1 gives x = , k = 2 gives x = . 18 9 π π 7π 4π 2 π 5 π 3 π + + + + + = . In conclusion, the sum of the roots is 18 9 18 9 9 18 2 2. If sin6 x = –

EXAMPLE

186 Solve 1 – sin x = sin ( 2x – 4π).

Solution

2

Let us use the sum and difference formulas to expand the right side: x π x π x π x 2 x 2 sin 2 ( – ) = (sin ⋅ cos – cos ⋅sin ) 2 = (sin ⋅ – cos ⋅ ) 2 2 4 2 4 2 4 2 2 2 2 =(

2 2 1 x x x x ) (sin – cos ) 2 = (sin – cos ) 2. 2 2 2 2 2 2

Now consider the left side. We can see that on the right side we have half angles, so let us use half angles on the left side too. x x x x We can use the identities sin x = 2 sin cos and 1= sin 2 + cos 2 to get 2 2 2 2 x x x x 2 2 x 2 x 1 – sin x = sin + cos – 2 sin cos = (sin – cos ) . 2 2 2 2 2 2 So the original equation becomes 1 x x x x x x x x (sin – cos ) 2 = (sin – cos ) 2,, i.e. 2(sin – cos ) 2 – (sin – cos ) 2 = 0. 2 2 2 2 2 2 2 2 2 x x 2 x x x x If (sin – cos ) = 0 then sin – cos = 0, i.e. sin = cos . 2 2 2 2 2 2 x sin x 2 = tan x =1. Dividing both sides by cos gives x 2 2 cos 2 π x π π We know arctan 1= , so the solution is = + kπ, i.e. x = + 2 kπ, k ∈ Z. 4 2 4 2 238

Algebra 10

EXAMPLE

187 Solve cos 2x + sin 2x = cos x + sin x.

Solution

Let us change the sine functions to cosine functions by using the reduction formula π π π cos( – x) = sin x . So sin 2 x = cos( – 2 x) and sin x = cos( – x). 2 2 2 π π Then the equation becomes cos 2 x + cos( – 2 x) = cos x+ cos( – x). 2 2

Using the sum to product formulas gives 2cos(

2x +

π π π π x+ – x x– + x – 2x 2x – + 2x 2 2 2 2 ) ⋅ cos( ) = 2cos( ) ⋅cos( ), i.e. 2 2 2 2

π π π π ⋅ cos(2 x – ) = 2 cos ⋅cos( x – ) . 4 4 4 4 π π π π Dividing by 2 cos on both sides ( cos ≠ 0 ) gives cos(2 x – ) = cos( x – ). 4 4 4 4 2 cos

There are two possible cases: 1. If 2 x –

π π = x – + 2 kπ then x = 2kπ, k ∈ . 4 4

π π π π 2n π = –( x – )+ 2 nπ then 3 x = + 2 n π and so x= + , n ∈Z. 4 4 2 6 3 π 2 nπ }, k, n ∈ Z. In summary, the answer is x ∈ {2 kπ, + 6 3

2. If 2 x –

Check Yourself 27 1. Solve |cos(45° – x)| =

2 . 2

2. For which values of m does the equation 2 – π 3. Solve arccos x – arcsin x = . 6 4. Solve each system of equations. ⎧ 3 ⎪sin x = – a. ⎨ 2 ⎪ ⎩cos x > 0

5. Solve the equations. a. cos2 3x + cos2 x = 1 c. cos(πcot x) = sin (πcot x) e. 43tan x – ñ3 = 1 Trigonometry: Functions, Equations and Inequalities

m = 2 cot x have a solution? sin x

1 ⎧ 2 2 ⎪⎪sin x + sin y = 2 b. ⎨ ⎪ x – y = 4π ⎪⎩ 3

b. sin4 x + cos4 x = sin x ⋅ cos x x d. cos x +1= cot 2 239

Answers 1.

kπ , k ∈Z 2

2. m ∈ [–2ñ2, 2ñ2]

π 4. a. – + 2kπ, k ∈ Z 3

1 2

π 7π b. { + kπ, – + kπ}, k ∈ Z 6 6

π kπ π πn 5. a. { + , + }, k, n ∈ Z 8 4 4 2

d. {π + 2kπ,

3.

π b. + kπ, k ∈ Z 4

π + 2 nπ}, k, n ∈ Z 2

1 c. {arccot + kπ}, k ∈ Z 4

e.

π + kπ, k ∈ Z 6

C. TRIGONOMETRIC INEQUALITIES 1. Basic Trigonometric Inequalities We can solve trigonometric inequalities by looking at the graphs of the basic trigonometric functions on the unit circle. By using the following steps we can find the solution of any simple trigonometric inequality: 1. Find the region which satisfies the given inequality on the unit circle. 2. Write the boundaries of the selected region. We calculate the boundaries by moving in an anticlockwise direction. Remember that the smaller bound (for example, a negative bound) must always be the first bound. 3. For sin x and cos x add 2kπ, and for tan x and cot x add kπ. To understand why step 3 is necessary, look at the figure. We know that trigonometric functions are periodic, and so a trigonometric equation or inequality has infinitely many solutions. To include all these answers we add kπ or 2kπ to the solution of any trigonometric inequality. y sin x >

1 1 2

1 2 x

–1

240

Algebra 10

EXAMPLE

188 Solve the inequality sin x > 21 .

Solution

Let us draw the unit circle and shade the required region. y

y

5p 6

1 2

y 1 2

x

p 6

1 2

5p 6

p 6 x

x

π 5π < x < . To get the full answer we must 6 6 π 5π add 2kπ to both sides, so the answer will be + 2 kπ < x < + 2 kπ, k ∈ Z. 6 6

As we can see in the figure, the boundaries are

a. Inequalities in sin x

y

Consider the inequality sin x > a. If a > 1 there is no solution (because –1 ≤ sin x ≤ 1).

p – arc sin a

a

arc sin a

If a < –1 the answer will be all the real numbers, i.e. (–∞, ∞).

x

If –1 ≤ a ≤ 1 the answer is (arcsin a + 2kπ) < x < (π – arcsin a + 2kπ), k ∈ Z. Remember! [a, b] is an interval which includes a and b: x ∈ [a, b] means a ≤ x ≤ b. (a, b) is an interval which does not include a or b: x ∈ (a, b) means a< x < b.

We can write this as y

x ∈ (arcsin a + 2kπ, π – arcsin a + 2kπ), k ∈ Z. Now consider the inequality sin x < a. If a < –1 there is no solution. If a > 1 the answer will be all the real numbers, i.e. (–∞, ∞).

–p – arc sin a

a

arc sin a

x

If –1 ≤ a ≤ 1 the answer is (–π – arcsin a + 2kπ) < x < (arcsin a + 2kπ), k ∈ Z.

Trigonometry: Functions, Equations and Inequalities

241

EXAMPLE

189 Solve the inequality sin x ≥

Solution

2 . 2

Let us draw the figure, as shown at the right. 2 π = , so the 2 4 π 3π second boundary is π – = . 4 4

We know arcsin

y 3p 4

ñ2 2

p 4

x

We have the ≥ sign, so the answer is 3π π ( + 2 kπ) ≤ x ≤ ( + 2 kπ), k ∈ . 4 4

We can write this answer alternatively as 3π ⎡π ⎤ x ∈ ⎢ + 2kπ , + 2kπ ⎥ , k ∈ . 4 4 ⎣ ⎦

EXAMPLE

190 Solve the inequality sin x > – 23 .

Solution

Let us draw the figure.

y

3 π ) = – , so the We know arcsin(– 2 3

π 4π second boundary is π – (– ) = . 3 3

x

So the answer will be 4π π (– + 2 kπ) < x < ( + 2 kπ), k ∈ , i.e. 3 3 4π π x ∈ (– + 2 kπ, + 2 k π), k ∈ . 3 3

EXAMPLE

ñ2 – 2

–p 3

191 Solve the inequality sin x < –2.

Solution 242

4p 3

We know that –1 ≤ sin x ≤ 1, so this inequality has no solution. Algebra 10

EXAMPLE

192 Solve the inequality 2sin 5x ≤ ñ2.

Solution

Rearranging the terms give us sin5 x ≤ We know arcsin

Remember! The first bound is always 5π the smallest: – is 4 π smaller than . 4

boundary, –π –

so

2 π = 2 4

the

is the second

first

boundary

y – 5p 4

p 4

ñ2 2

is

π 5π =– . 4 4

x

By solving the equation we get (−

5π π + 2 kπ) ≤ 5 x ≤ ( + 2 kπ), 4 4

k∈ .

Dividing all sides by 5 gives us ( −

EXAMPLE

2 . 2

2kπ π 2kπ π + ) ≤ x ≤( + ), k ∈ . 4 5 20 5

193 Solve the inequality 4sin3x + 2 < 3.

Solution

1 Rearranging the terms give us sin 3x < . 4 1 arcsin is not a special angle so, we will leave it 4

y

1 4

– p – arcsin 1 4

in this form.

arcsin 1 4

So we can write the answer as ( −π − arcsin

x

1 1 + 2 kπ)<3 x< (arcsin + 2 k π) , k ∈ . 4 4

Dividing all sides by 3 gives us (−

π + arcsin 3

1 1 arcsin 4 + 2kπ ) < x < ( 4 + 2k π ), k ∈Z. 3 3 3

b. Inequalities in cos x Consider the inequality cos x > a.

y arccos a

If a > 1 there is no solution. If a < –1 the answer will be all the real numbers, i.e. (–∞, ∞).

a x

If –1 ≤ a ≤ 1 the solution is (–arccos a + 2kπ) < x < (arccos a + 2kπ), k ∈ .

Trigonometry: Functions, Equations and Inequalities

– arccos a

243

Now consider the inequality cos x < a. If a > –1 there is no solution.

y arccos a

If a < 1 the answer will be all the real numbers, i.e. (–∞, ∞).

a

If –1 ≤ a ≤ 1 the answer is

x

arccos a + 2 kπ < x < 2 π – arccos a + 2 kπ, k∈ .

EXAMPLE

2p – arccos a

194 Solve the inequality cos x ≤ 0.

Solution

Let us draw the figure, as shown at the right. The boundaries are arccos0 = 2π –

π 3π = . So 2 2

y

π and 2

p 2

x

3π π ( + 2 kπ) ≤ x ≤ ( + 2 kπ), k ∈ . 2 2 3p 2

EXAMPLE

195 Solve the inequality 2cos 2x – 1 > 0.

Solution

1 Rearranging the terms gives us cos 2 x > . 2 1 π The boundaries are arccos = and 2 3 1 π − arccos = − . So we have 2 3

y

p 3

1 2

π π ( − + 2kπ) < 2 x < ( + 2 k π), k ∈ Z. 3 3

Dividing all sides by 2 gives π π ( − + kπ) < x < ( + k π), k ∈ Z. 6 6 244

x

–p 3

Algebra 10

EXAMPLE

196 Solve the inequality 2cos x + ñ3 ≤ 0 in [0, 2π].

Solution

3 . 2 Now let us draw the figure. The boundaries

Rearranging the terms gives us cos x ≤ –

are arccos( − 2π − arccos( −

5p 6

3 5π and )= 2 6

– ñ2 2

3 5π 7 π )= 2π − = . 2 6 6

So the answer is [

EXAMPLE

y

x

5π 7 π , ]. 6 6

7p 6

197 Solve the inequality 2cos 3x > –π.

Solution

π π We can rewrite the inequality as cos 3 x > − . However, π ≅ 3.14 and so − < −1. So the 2 2 inequality is true for all real numbers: x ∈ (–∞, ∞).

Check Yourself 28 1. Solve the inequalities. a. sin x <

2 2

b. 2sin x ≤ –ñ3

π d. sin(5 x + ) ≥ 0 3

c. 2sin 3x < 1

e. 2sin (–2x) > 1

2. Solve the inequalities. a. 2cos x ≤ ñ2 d. cos( −2 x) >

b. 2cos(3x – 2 2

π )≥1 6

c. cos 3 x < –

3 2

e. 5cos 3x – 2 > 0

3. Solve the inequalities in the interval [0, 2π]. π 2 a. cos(2 x − ) ≥ 4 2

π b. sin(3 x + ) < 0 3

Answers 5π π 1. a. − + 2kπ < x < + 2k π, k ∈ 4 4 7π 2 kπ π 2kπ + < x< + , k∈ 18 3 18 3 5π π e. − + kπ < x < − + k π, k ∈ 12 12

c. −

Trigonometry: Functions, Equations and Inequalities

c. 3sin x + 1 ≥ 0

b. −

2π π + 2kπ ≤ x ≤ − + 2kπ, k ∈ 3 3

d. −

2 π 2 kπ π 2 kπ + + ,k∈ ≤x≤ 15 5 15 5

245

2. a. c.

7π π + 2kπ ≤ x ≤ + 2kπ, k ∈ 4 4

b. −

π 2kπ π 2kπ + , k∈ ≤x≤ + 18 3 6 3

5π 2 kπ 7 π 2kπ + < x< + , k∈ 18 3 18 3

π π d. − + kπ < x < + kπ, k ∈ 8 8

2 2 − arccos + 2 kπ arccos + 2 k π 5 5 e. <x< , k∈ 3 3

2π 5π 8 π 11π 14 π 17 π ⎡ π ⎤ ⎡ 5π ⎤ 3. a. ⎢0, ⎥ ∪ ⎢π, b. ( , ) ∪ ( , ) ∪( , ) ⎥ 9 9 9 9 9 9 4 ⎦ ⎣ 4⎦ ⎣ 1 1 c. [0, π + arcsin ] ∪[2 π – arcsin , 2 π] 3 3

c. Inequalities in tan x y

Consider the inequality tan x > a.

p 2

As we can see in the figure, the solutions to the inequality are symmetric about the origin. So we will write only one answer and add kπ instead of 2kπ to give the final solu-

arc tan a x

tion. So the solution is

arctana + kπ < x <

π + kπ , k ∈ . 2

p + arc tan a

y

Now consider the inequality tan x < a. From the figure we can get the solution −

π + kπ < x < (arctan a + k π), k ∈ 2

–

p 2

p 2

.

a arc tan a x

p + arc tan a –

p 2

Note π If the inequality is given with ≥ or ≤, the answer will include arctan a but exclude ± , 2 π because tan( ± ) is undefined. 2 246

Algebra 10

EXAMPLE

198 Solve the inequality tan x > 1.

Solution

y

Look at the figure. π arctan1= , so the answer is 4

p 2

1 p 4

π π ( + kπ) < x < ( + kπ), k ∈ . 4 2

x

–

EXAMPLE

p 2

tan x

199 Solve the inequality tanx ≤ ñ3.

Solution

y

Look at the figure. We know arctan 3 =

ñ3

p 2

π , so the answer is 3

p 3

π π ( − + kπ) < x ≤ ( + k π), k ∈ . 2 3

x

Alternatively, we can write π ⎛ π ⎤ x ∈ ⎜ − + kπ, + kπ ⎥ , k ∈ . 3 ⎦ ⎝ 2

EXAMPLE

200 Solve the inequality 3tan(4x – 5π )+

Solution

–

π 3 tan(4 x − ) ≥ − . We know 5 3 3 π arctan(– ) = – , so we have 3 6 π π π π to (− + kπ) ≤ 4x − < ( + kπ) . Adding 5 6 5 2 π 7π each part gives ( + kπ) ≤ 4x < ( + k π). 30 10

(

tan x

3 ≥ 0.

Rearranging the terms gives us

So the final answer is

p 2

y

p 2

x

p – 2

–

p 6

–

ñ3 3

tan x

7 π kπ π kπ + ) ≤ x < ( + ), k ∈ . 120 4 40 4

Trigonometry: Functions, Equations and Inequalities

247

EXAMPLE

201 Solve the inequality ñ3tan x ≥ 1 in [0, 2π].

Solution

3 . 3 3 π π Since arctan = , the boundaries are 3 6 6

y

Rearranging the terms gives us tan x ≥

p 2 ñ3 3

π 7π and π + = . So we have 6 6 ⎡ π π ⎞ ⎡ 7 π 3π ⎞ x∈ ⎢ , ⎟ ∪ ⎢ , ⎟. ⎣6 2 ⎠ ⎣ 6 2 ⎠

p 6 x

7p 6 3p 2

d. Inequalities in cot x

tan x

y

cot x

a

Consider the inequality cot x > a and look at the figure. We can see that the inequality has solution kπ < x < (arccot a + kπ), k ∈ .

arccot a

x

p + arccot a

Now consider cot x < a.

y

From the figure we can get the solution

a

(arccot a + kπ) < x < (π + kπ), k ∈ .

cot x

arccot a

x

p + arccot a

Note If the inequality is given with ≥ or ≤, the answer will include arccot a but exclude 0 and π, because cot 0 and cot π are undefined. 248

Algebra 10

EXAMPLE

202 Solve the inequality cot x > 1.

Solution

EXAMPLE

π π We know arccot 1= , so by the formula we have kπ < x < ( + kπ), k ∈ . 4 4

203 Solve the inequality 3cot 2x + ñ3 ≤ 0.

Solution

Rearranging the terms gives us cot 2 x ≤ – We know arccot(–

3 . 3

3 2π )= , so we have 3 3

2π ( + kπ) ≤ (2 x < π + kπ) . Dividing all parts by 3 2 gives us the answer:

–

ñ3 3

y cot x

2p 3 x

π kπ π kπ ( + ) ≤ x < ( + ), k ∈ . 3 2 2 2

EXAMPLE

204 Find all solutions of the inequality cot 2x ≥ –ñ3 in [0, 2π].

Solution

y

Let us draw the figure. From the figure we have kπ < 2 x ≤

5π + kπ 6

5π kπ kπ and so < x ≤ ( + ), k ∈ . 2 12 2

5p 6 x

p

If we replace k with 0, 1, 2 and 3 we get 11p 6

5π ⎤ ⎛ π 11π ⎤ ⎛ 17 π ⎤ ⎛ 3 π 23 π ⎤ ⎛ x ∈ ⎜ 0, ⎥ ∪⎜ , ⎥ ∪ ⎜ π, 12 ⎥ ∪ ⎜ 2 , 12 ⎥ . ⎝ 12 ⎦ ⎝ 2 12 ⎦ ⎝ ⎦ ⎝ ⎦

If k is greater than 3 or smaller than 0 then the answer is not in [0, 2π], so this is the complete solution. Trigonometry: Functions, Equations and Inequalities

249

Check Yourself 29 1. Solve the inequalities. a. tan 3 x <

3 3

π b. tan( x – ) – 3 ≥ 0 3

c. 2tan 3x < 6

2. Solve the inequalities. a. cot 3x ≤ –1

b. ñ3cot(2x +

π )≥1 6

c. 2cot 4x – 1 < 0

3. Solve the inequalities in the interval [0, π]. π a. tan(3 x + ) ≥ − 3 3

π b. cot(2 x – ) <1 3

Answers π kπ π kπ 1. a. − + < x< + , k ∈ 6 3 18 3

c. –

b.

2π 5π + kπ ≤ x < + kπ, k ∈ 3 6

π kπ 1 kπ + < x < arctan3 + , k ∈ Z 6 3 3 3

π kπ π kπ + ≤ x< + , k ∈ 4 3 3 3 1 arccot π kπ 2 + kπ , k ∈ c. + < x< 4 4 4 4

2. a.

b. −

π kπ π kπ + <x≤ + , k∈ 12 2 12 2

π ⎞ ⎡ π 7 π ⎞ ⎡ 4 π 13 π ⎞ ⎡7 π ⎤ ⎡ 3. a. ⎢0, , π⎥ ⎟∪⎢ , ⎟∪⎢ , ⎟∪ 18 ⎠ ⎣⎢ 9 ⎣ 18 ⎠ ⎣ 9 18 ⎠ ⎣ 9 ⎦ ⎡ π ⎞ ⎛ 7 π 2 π ⎞ ⎛ 19 π ⎤ b. ⎢0, ⎟ ∪ ⎜ , , π⎥ ⎟∪⎜ ⎣ 6 ⎠ ⎝ 24 3 ⎠ ⎝ 24 ⎦

2. Advanced Trigonometric Inequalities Some problems may ask us to solve more complex trigonometric inequalities. The inequality may include one type of trigonometric ratio but with different powers, or it may include two or more different types of ratio. Alternatively, the expression may include one type of trigonometric ratio with two boundaries. 1. If the question has two inequalities (for example, a < cos x < b), we can draw a unit circle and find the answer on the circle by using the methods we have already seen. 2. If the inequality contains a trigonometric ratio with different powers, we can try to factorize it. By using substitutions we can change the inequality into a quadratic inequality and then solve it. We can then use this to write the final solution. 250

Algebra 10

3. If the problem involves more than one type of trigonometric ratio, we can try to simplify or factorize it by using the methods that we have learned in the previous sections. We can use summation formulas, sum to product or product to sum formulas, trigonometric identities or other types of substitution to simplify the expression. 4. If we have a system of trigonometric inequalities, we can try to solve each inequality separately and then find their common solution. For some questions we may have to use a combination of methods to obtain the solution. Let us look at some examples.

EXAMPLE

205 Solve the inequality − 21 < cos x ≤

Solution

2 . 2

Look at the figure. We can use 2 π 1 2π = and arccos( − ) = , and 2 4 2 3 π 2π find the other boundaries as – and − . 4 3

2p 3

arccos

–

y

1 2

p 4

x

ñ2 2

Now we are ready to write the solution as the union of the two regions in the figure:

–

2p 3

–p 4

2π π ⎡π ⎞ ⎛ 2π ⎤ ⎢ 4 + 2kπ, 3 + 2kπ ⎟ ∪ ⎜ − 3 + 2 nπ, − 4 + 2 nπ ⎥ , k, n ∈ . ⎣ ⎦ ⎠ ⎝

EXAMPLE

206 Solve the inequality

Solution

3 1 > sin x > . 2 2

Let us draw the figure. Using arcsin

3 π 1 π = and arcsin = , we 2 6 2 3

5p 6

2p 3

y

p 3

p 6

can find the other boundaries as π 5π π 2π and π − = π− = . 6 6 3 3

x

The answer is the union of the intervals: π π 2π 5π ( + 2 kπ, + 2 kπ) ∪ ( + 2 n π, + 2 n π), k, n ∈ . 6 3 3 6 Trigonometry: Functions, Equations and Inequalities

251

EXAMPLE

207 Solve the inequality |tan x| ≤ 1.

Solution

We can rewrite this inequality as

y

p 4

3p 4

–1 ≤ tan x ≤ 1.

1

Let us draw the figure. We have arctan1=

π π and arctan( −1) = − , so the 4 4

x

π π answer is [− + kπ, + kπ], k ∈ . 4 4

EXAMPLE

–

3p 4

–

p 4

–1

208 Solve the inequality 2sin x + sin x – 1 ≥ 0. 2

Solution

First notice that the inequality contains a quadratic expression in sin x. Let us use the substitution t = sin x and try to factorize the new expression. The inequality becomes 2t2 + t – 1 ≥ 0, which we can factorize as (2t – 1)(t + 1) ≥ 0. The roots of this expression are 1 t = and t = –1. Let us draw a table for the inequality: 2 1 2

–1 2t2 + t – 1

+

–

+

So the inequality is true for sin x ≤ –1 and sin x ≥

1 . 2

π If sin x ≤ –1 then x = − + 2 kπ, k ∈ . (1) 2 1 5π π If sin x ≥ then + 2 n π ≤ x ≤ + 2 n π, n ∈ . (2) 2 6 6 π π 5π (1) and (2) give us the solution: x = − + 2 kπ, k ∈ Z and + 2 n π ≤ x ≤ + 2 n π, n ∈ Z. 2 6 6 EXAMPLE

209 Solve the inequality 1 – 4cos x < 0.

Solution

2

Let us use the substitution t = cos x and try to factorize the expression: 1 1 1 – 4t2 = (1 – 2t)(1 + 2t) < 0. The roots are t = and t = − and the inequality is true for 2 2 1 1 t < – and t > . 2 2

252

Algebra 10

1 2π 4π So the solutions are cos x < − , i.e. + 2 kπ < x < + 2 k π, k ∈ , and 2 3 3 1 π π cos x > , i.e. − + 2 nπ < x < + 2 n π, n ∈ . 2 3 3 4π π ⎛ 2π ⎞ ⎛ π ⎞ Finally, the answer is x ∈ ⎜ + 2kπ, + 2 k π ⎟ ∪ ⎜ − + 2 n π, + 2 n π ⎟, n, k ∈ . 3 3 ⎝ 3 ⎠ ⎝ 3 ⎠ EXAMPLE

210 Solve the inequality sin5x cos 3x − cos5 xsin 3 x<

Solution

3 . 2

We can use the sum and difference formulas on the left-hand side: sin 5x cos 3x – cos 5x sin 3x = sin (5x – 3x) = sin 2x. 3 . From the 2 3 π figure, the second boundary is arcsin = 2 3 π 4π and the first boundary is −π − = − . 3 3

So we need to solve sin 2 x <

y –

4p 3

ñ3 2

p 3

x

So we have π 4π + 2kπ < 2 x < + 2 k π, i.e. 3 3 π 2π ( − + kπ) < x < ( + k π), k ∈ . 3 6 −

EXAMPLE

211 Solve the inequality (cos 2x – 3cos x) ≥ 1.

Solution

It is difficult to work with the different ratios cos 2x and cos x, so let us use the formula cos 2x = 2cos2x – 1 to get an inequality in cos x only. Then take 1 to the left side of the inequality to get 2cos2x – 3cos x – 2 ≥ 0. Using the substitution t = cos x gives us 2t2 – 3t – 2 ≥ 0. By factorizing we have (2t + 1)(t – 2) ≥ 0. The roots are t = − – 2

2t – 3t – 2

+

1 2

1 and t = 2. Write a table: 2

2 –

+

1 or cos x ≥ 2. The second inequality has no solution, so we only 2 1 2π 4π need to consider cos x ≤ – ⇒ ( + 2 kπ) ≤ x ≤ ( + 2 kπ), k ∈ . 3 3 2

So the solution is cos x ≤ –

Trigonometry: Functions, Equations and Inequalities

253

EXAMPLE

212 Solve the inequality (sin x – cos x) ≥ sin 2x. 2

Solution

Expanding the power gives us sin2x – 2sin x cos x + cos2x ≥ sin 2x. Using the identities sin2x + cos2x = 1 and 2sin x cos x = sin 2x, we can rewrite this 1 inequality as 1 – sin 2x ≥ sin 2x, i.e. 1 ≥ 2sin 2x and sin 2 x ≤ . 2 1 1 π π 7π are arcsin = and − π − = − . 2 2 6 6 6 7π 7π π π So the solution is − + 2kπ ≤ 2 x ≤ + 2 k π, i.e. ( − + k π) ≤ x ≤ ( + k π), k ∈ . 6 6 12 12

The boundaries for sin 2 x ≤

EXAMPLE

⎧sin x > 0

213 Solve the system of inequalities ⎨⎩cos x ≤ 0 .

Solution

Let us draw the unit circle and find the answers on it.

y

y

In the first unit circle the answer is (0, π), π 3π In the second unit circle the answer is [ , ] 2 2 The intersection of these two answers gives us π the solution: x ∈ [ + 2kπ, π + 2 kπ), k ∈ . 2 EXAMPLE

x

sin x > 0

cos x £ 0

⎪⎧tan x ≥ 3 . ⎨ ⎪⎩cot x <1

214 Solve the system of inequalities

Solution

x

Let us draw the unit circle and find the answers on it. y p 2

ñ3

y

p 6

p 4

x

p

tan x ³ ñ3

π π In the first unit circle the answer is [ + kπ, +kπ), k ∈ . 3 2 254

1

x

cot x < 1

(1) Algebra 10

π In the second unit circle the answer is ( + πn, π + πn), n ∈ . (2) 4 π π The intersection of (1) and (2) gives us the solution: x ∈ [ + kπ , + kπ), k ∈ . 3 2

Check Yourself 30 1. Solve the inequalities. a. −

3 1 < sin x ≤ 2 2

d. cos 2 x <

1 4

g. cos 2 2 x < sin 2 2 x+

b. |tan x| > ñ3

c. 8sin2x – 6sin x + 1 ≥ 0

e. 4tan2x – tan x – 3 < 0

f.

6 < 3+ 2 cos x cos x +1

2 2

2. Solve the inequality sin x + cos x > 0 in [0, 2π]. ⎧ ⎪⎪sin x < 3. Solve the system of inequalities ⎨ ⎪ cos x ≥ ⎪⎩

2 2 . 1 2

Answers π 4π ⎛ π ⎤ ⎡ 5π ⎞ 1. a. ⎜ − + 2kπ, + 2kπ ⎥ ∪ ⎢ + 2 nπ, + 2 nπ ⎟ , n, k ∈ 3 6 6 3 ⎦ ⎣ ⎝ ⎠ π π π π b. ( − + kπ, − + kπ) ∪ ( + n π, + n π), n, k ∈ 2 3 3 2

1 1 π 5π c. [− π − arcsin + 2kπ, arcsin + 2kπ] ∪ [ + 2 nπ, + 2 nπ], n, k ∈ 4 4 6 6 π 2π d. ( + kπ, + kπ), k ∈ 3 3

3 π e. − arctan + kπ < x < + k π, k ∈ 4 4

π π f. ( − + 2kπ, + 2 kπ), k ∈ 3 3

g.

2. [0,

3π 7π ) ∪ ( , 2 π] 4 4

Trigonometry: Functions, Equations and Inequalities

7 π kπ π kπ + < x< + , k∈ 16 2 16 2

π ⎡ π ⎞ 3. ⎢ − + 2kπ, + 2kπ ⎟ , k ∈ 3 4 ⎣ ⎠ 255

ABUL WAFA BUZJANI (10 June 940 - 15 July 998)

Abul Wafa Buzjani was a Muslim mathematician and astronomer. He was born in Buzjan, Khorasan in Iran. His extended name is Abu al-Wafa Muhammad ibn Muhammad ibn Yahya ibn Ismail ibn al-Abbas al-Buzjani. When he was nineteen years old, Abul Wafa moved to Iraq. He made many discoveries in mathematics and astronomy. Abul Wafa worked mostly on trigonometry and wrote many books, but many of them are lost today. Abul Wafa also studied the movement of the moon. There is a crater on the moon with his name. He also invented a wall quadrant for the calculation of measurements in astronomy. Abul Wafa is known as the father of the basic rules of plane trigonometry and spherical trigonometry. He introduced the terms ‘secant’ and ‘cosecant’ to mathematics and improved methods for calculating trigonometric tables. He also developed new methods for solving some problems in spherical triangles. In spherical trigonometry, he studied and developed the properties of the sine function. He also established some of the basic trigonometric identities and formulas, including sin(a + b) = sin a ⋅ cos b + cos a ⋅ sin b sin 2a = 2sin a ⋅ cos a cos 2a = 1 – 2sin2a sin2a = 1 – cos2a. Abul Wafa discovered the law of sines for spherical triangles that sin A sin B sin C is = = . sin a sin b sin c The works of Abul Wafa were translated to Western languages beginning

in

the

12th

century.

He

is

a

well-known

mathematician, and many books have been written about him. Abul Wafa died on 15 July 998 in Baghdad, Iraq.

Spherical triangles

NASIR AL-DIN AL-TUSI (18 Feb 1201 - 26 June 1274) Nasir al-Din al-Tusi was born in 1201 in Tus, Khorasan in Iran. His extended name is Muhammad ibn Muhammad ibn al Hasan al-Tusi, but he is known as just Tusi in the West. Tusi was a Muslim mathematician, astronomer, chemist, biologist, physicist and scientist. He also studied Islamic theology and philosophy. Tusi began his studies at an early age. In Tus he studied the Koran, hadith, logic, philosophy, mathematics, astronomy and medicine. Later he traveled to different places to study and to attend the lectures of famous scholars. He went to Nishapur to study philosophy and mathematics. After the attack of the Mongols he went to Mosul and other places, and during this time he made his most important contributions to science. Finally he joined Hulagu Khan’s army. Tusi died in 26 June 1274 in Kadhimiya, Iran, and left about 150 works on different subjects. A 60 – kilometer diameter crater on the moon was named Nasireddin, in honor of Tusi. The minor planet 10269 Tusi was also discovered by a Russian astronomer and given his name. Tusi was the first scientist to treat trigonometry as a separate mathematical discipline, different from astronomy. He also gave the first extensive description of spherical trigonometry, and he was the first person to list the six distinct cases of a right triangle in spherical trigonometry. Tusi formulated the law of sines for plane triangles: a b c = = . He also wrote the same formula for sin A sin B sin C spherical triangles and proved these laws. In 1265, Tusi wrote a manuscript regarding the calculation for the nth roots of an integer. He studied binomial expansions, and used Pascal’s triangle before Pascal. Today Tusi’s work lives on in his many books about mathematics, biology, astronomy and chemistry.

EXERCISES

3 .4

A. Types of Trigonometric Equation 1. Solve the equations. a. sin x =

2 2

3. Solve the equations. a. tan x = –1

b. sin x = −

3 2

b. 3tan 2x – ñ3 = 0 2π ) −2=0 3

c. 6sin 2x + 3 = 0

d. 2sin 3x – ñ3 = 0

c. 2 tan(5 x −

π e. 2 sin( x − )+ 3 = 0 5

f. 4sin 5x = 0

d. 3tan(3x + 24°) – ñ3 = 0

x π g. sin( − ) = −1 3 4 3π h. 2 sin( + 2 x) − 1= 0 i. 3sin 4x + 2 = 0 2 x j. sin( π − ) − 1= 0 2 x 2 sin 2 x k. 2 sin( − 2 π)+ 2 = 0 l. =1 3 2

x π e. tan( + ) = 2 3 2 3π f. tan( x − ) = 0 2

4. Solve the equations. a. cot x = 1 b. cot 4x = ñ3

2. Solve the equations. a. cos x =

2 2

π 2 c. cos(5 x − ) = 4 2

b. 2cos x + ñ3 = 0 d. 3cos 4x – 4 = 0

e. 2cos 3x – ñ3 = 0 f.

2π cos(2 x + ) = −1 3

g. cos(5x – 30°) = 1 π 2 h. cos( − 2 x) = in [0, 2 π] 2 2 3π x i. cos( + ) = 0 in [0, 2 π] 2 3 258

c. 1 – cot 2x = 2 π d. 3cot(3 x − )+ 3 = 0 3 π x e. cot( − ) = −1 2 3 x f. 2 cot( − ) − 4 = 0 2

5. Find the sum of the roots of each equation in the interval 700° < x < 1000° in degrees. a. cos 3 x =

1 2

b. sin 2 x =

2 2

π c. tan( x + ) = 3 3 Algebra 10

6. Solve the equations.

9. Solve the equations.

a. cos 3x + cos 5x = 0

a. sin 3x = sin 30°

b. 4cos22x + 2cos 2x = 0

b. cos 4x = cos 20°

2

c. 2cos x + cos x = 1 3cos x d. cos 2 x +1= 2 e. cos 2x = sin x + 1

c. tan x = tan d. cot

f. cos 3x + cos 5x = 2cos 4x

10. Solve

π 4

x = cot15 ° 2

sin 3x cos 3 x . = sin x cos x

7. Solve the equations. a. sin 2x – sin x = 0 b. (sin 3 x cos 2 x)+(cos 3 xsin 2 x) =

1 2

π 4

π 3

11. Solve tan(2 x + ) = cot( x − ) .

c. sin 6x = sin 2x d. 2sin 3x = 1 e. cos 2x = 3cos x + 1 in [0, 2π] f.

2 sin x − 2 = 0 in [0, 2 π] 2 sin x + 3

g. sin(2x + 10°) – sin(2x – 10°) = 0 3

2

h. 4sin x + 2sin x – 2sin x = 1

12. Solve each linear equation in sin x and cos x. a. 3sin x – 3cos x = 5 b. 3cos x – 4sin x = 5 c. 12sin x + 5cos x = 13 d. ñ2sin x + ñ2cos x = 1 e. ñ3sin x – cos x = 1

8. Solve the equations. a.

›f. sin x – 2cos x = 1

2 tan 3 x = 3 1 − tan 2 3x

b. tan2x + tan x – 2 = 0 c. cot2x – 1 = 0 cot 3x − cot x =− 3 d. 1+ cot 3 x cot x

e. tan x + cot x = 2 f.

x cot − cos x =1 2

Trigonometry: Functions, Equations and Inequalities

13. Solve the homogeneous equations. a. 2sin x + cos x = 0 b. 3sin x – ñ3cos x = 0 c. sin2x + sin x cos x – 2cos2x = 0 d. sin2x – 2sin x cos x – 3cos2x = 0 259

14. Find the maximum and minimum values of each

20. For which values of a does the equation ›

function.

5−

a. f(x) = 3sin x + 4cos x

a = 3tan x have a solution? cos x

b. f(x) = ñ3sin x + cos x c. f(x) = 2sin x – ñ5cos x d. f(x) = sin x – cos x

π 2

21. Solve arccos x+ arctan x = . ›

15. Find the range of each function. a. f(x) = 5sin x – 12cos x + 13 b. f(x) = ñ7sin x + ñ2cos x + 1 c. f(x) = sin x + ñ3cos x – 3

π ⎧ ⎪x + y = 4 Solve the system of equations. 22. ⎨ › ⎪ ⎩tan x + tan y =1

B. Further Trigonometric Equations 16. Solve |sin x – cos x| = sin x + cos x.

5 2

23. Solve tan2 x +cos 2 x = . ›

17. Solve each system of equations in the interval [0, π]. 1 ⎧ ⎪sin x sin y = 4 a. ⎪⎨ π ⎪ ⎪⎩ x + y = 3

24. Solve tan2x + cot2y – 2tan x – 4cot y + 5 = 0. ⎧sin( x + y) =1 b. ⎪⎨ ⎪⎩sin( x − y) =1

⎧tan x + cot y =1 c. ⎪ ⎨ π ⎪x + y = 4 ⎩

18. Find the roots of sin4 x + cos4 x = cos 4x in [0, 2π].

19. Solve 1 – cos 2x = 2 sin x. 260

›

C. Trigonometric Inequalities 25. Solve each inequality. 1 2 π b. 2 sin( x − )+1< 0 3

a. sin 3x > −

c. 2sin 2x + ñ2 ≥ 0 π 3 d. sin(3 x − ) ≤ 5 2

e. 3sin 5x + 3 > 0 Algebra 10

26. Solve the trigonometric inequalities. a. cos 2 x > −

1 2

x c. 2 cos + 2 ≥ 0 3 x π e. 2 cos( − ) − 3 > 0 2 3

31. Solve the trigonometric inequalities.

π b. 2 cos(2 x − ) < −1 3

d. 3cos x – 3 > 0

a. tan2x < 3 b. tan2x + tan x – 2 > 0 c. (cot2x – 1)(cot2x – 3) ≤ 0

32. Solve the trigonometric inequalities.

27. Solve the trigonometric inequalities. a. tan 3x ≤ 1

b. 3tan 4x + ñ3 < 0

x 5π c. 3tan( − )+3 ≥ 0 2 36

d. tan 5x + ñ3 > 0

a. cos 4 x cos x + sin 4 xsin x ≥ 2 b. 1+ sin

x x ≥ 4sin 3 x − cos 2 3 3

c. 2(sin x + cos x)2 ≥ 3 d. sin x cos x <

28. Solve the trigonometric inequalities. a. cot 3x + ñ3 < 0

1 2 ≤ sin x < 2 2 3 c. |sin x| ≥ 2

1 4

e. cos 2x + sin x – 1 < 0

x b. 2 cot + 2 ≥ 0 2

29. Solve the trigonometric inequalities. a. −

2 2

b.

33. Solve the trigonometric inequalities.

1 3 < cos x ≤ 2 2

d. − 3 < tan x ≤

a. 3 3

6 tan 3 x > 3 tan 2 3x − 1

b. tan x – cot x > 0

e. |cot x| > 1

34. Solve each system of inequalities. 30. Solve the trigonometric inequalities. a. sin x – 2sin x ≥ 0 2

b. 2cos2x – cos x – 1 > 0 c. 6cos2x + cos x – 1 < 0 d. 2sin2x – 1 ≤ 0 e. 2cos4x – 3cos2x > –1 Trigonometry: Functions, Equations and Inequalities

1 ⎧ ⎪sin x cos x ≤ 2 a. ⎨ ⎪ ⎩sin x + cos x ≥ 2

⎧⎪tan x >1 b. ⎨ ⎪⎩cot x ≥ −1

1 ⎧ ⎪sin x cos x > 4 c. ⎨ ⎪ ⎩tan x ≥ 3 261

CHAPTER REVIEW TEST

3A

1. (a – 3)x2 + (b + 1)y2 = 1 is the equation of a unit circle. Find (a, b). A) (2, 1)

B) (1, 2)

D) (3, 4)

C) (3, 2)

5. This figure contains

E a D

seven identical squares. What is tan (∠EDB)?

B

C

E) (4, 0) A

A)

3 4

B)

4 3

C)

5 4

D)

3 5

E)

5 3

⎛ 3 ⎞ , b ⎟⎟ is a point on the unit circle. Given that ⎝ 2 ⎠

2. P ⎜⎜

6. In the figure,

P is in the fourth quadrant, find b.

A) –

1 2

B) –

3 2

C) –1

D)

1 2

E)

3 2

A

BC = ò10 and

x

1 tan α = . 3 Find AC = x.

B

A) 3ñ3

C) ñ3

a

B) 3

ò10

D)

3. sin x – cos x = 1 is given. What is sin x ⋅ cos x?

C

1

E) 1

3

4

A)

15 16

B)

15 32

C)

3 4

D)

3 8

E)

7 8

7. In the figure,

A

m(∠ABC) = 30°, m(∠BCA) = 90° and DB = DC. What is tan (∠DAC)?

a

30° B

4. In the figure, ABCD is a square with m(∠CKE) = 90°, m(∠DHA) = 90°, DH = HK and m(∠DAH) = α. Find tan α. 7 A) 2 262

3 B) 2

D

C H K

3 2

B)

5 2

C)

5 3

C

D) ñ3

E) 2ñ3

E

a

8. Which one of the following is equivalent to

A

3 C) 4

A)

D

B

2 D) 3

1 E) 2

sin 40°? A) sin 220° D) sin (–40°)

B) cos 140°

C) sin 50°

E) cos (–50°) Algebra 10

9. Which one of the following is not equivalent to ⎛π ⎞ sin ⎜ − α ⎟ ? 2 ⎝ ⎠ ⎛π ⎞ A) sin ⎜ 2 + α ⎟ ⎝ ⎠

B) cos (2π – α)

D) cos α

C) cos (–α)

E) sin (–α)

13. In the figure,

A

ABC is a triangle with AB = 10 cm, DE = BE = 5 cm, EC = 3 cm and DC = 4 cm. Find AD = x.

x D 5

B

A) 2

10.

3 4 = is given. Find the positive value of cos x sin x

cos x. 2 3

A)

B)

2 5

C)

3 5

D)

4 5

3 5

E)

B) 4

5

C) 6

D) 8

x

B

135°

B) 135°

C

ò10

C) 120°

A) 1

B) ñ2

C) 2

D) 2ñ2

E) 4

A)

D) 90°

6

A 2 C x

D

A) 5

B) 6

Chapter Review Test 3A

B

C) 8

B

B) 12 cm2

72 cm 2 5

C) 18 cm2

E) 15 cm2

16. An inscribed triangle has side lengths of 3, 7 and 8 units respectively. Find the circumference of its circumscribed circle.

5

A) 7ñ3π units

E

D) 10

E

9

8

4 12

E) 60°

5

6

C

54 cm 2 5

D)

12. Find x in the figure.

A

CD = 5 cm. Find A(ΔCDE).

Find x in the figure.

E) 10

D

AB ⊥ AE, AB = 8 cm, AC = 6 cm, CE = 9 cm and

ñ2

C

2 a = 5 cm and c = 6 cm. Which one of the following is a possible value of m(∠B)?

15. In the figure, A

3

14. ABC is a triangle such that A( ΔABC ) = 15 2 cm 2 ,

A) 150°

11.

E

4

E) 12

D)

B)

7 π units 3

7 3 π units 7

E)

C)

14 3 π units 3

3 3 π units 7 263

3B

CHAPTER REVIEW TEST 1. Find m(∠B) = α

5. In the figure,

A

in the figure. 23°41¢57¢¢

G F

ABCDEFGH is a cube. Find tan (∠GBH).

E

H

52°15¢23¢¢

a B

D

C

A) 28° 34′ 26′′

B) 27° 33′ 26′′

C) 27° 34′ 26′′

D) 28° 33′ 26′′

D

a

C

A

A)

1 3

2

B)

3

B

1

C)

3

D)

2

E) ñ3

2

E) 29° 33′ 26′′

6. The figure contains

2. Find the primary directed angle of –3333°. A) 277°

B) 267°

3. In the figure, ABCD

C) 263°

D) 257°

E) 253°

D

A

5 A) 12

4 B) 15

A)

x

5

α and β are b

C) 12 5

15 D) 4

20 E) 3

3 2

C)

I

H

D

E

F

G

1 5

D)

2 5

3 5

E)

2 tan θ – sec θ ? 15 B) 3

4

15 15 15 15 C) D) − E) − 3 5 5 15

8. π < β < π and sin 2 β = 12 are given. What is B

264

B)

J

K

A

3 and cos B = . 5 Find cot C.

1 2

B

7. π < θ < 3π and sin θ = – 1 are given. What is

4. In the figure, AB = BC

B)

L

C

A)

A) 2

1 2

a B

M

C

is a trapezoid and complementary angles, BD = 5 4 and tan α = . 3 Find DC = x.

five identical squares. What is sin (∠ALM)?

A

C)

3 4

C

D)

4 3

E)

5 4

4 2 tan β + cot β ?

A)

1 13

B)

1 13

13

C)

24 13

D)

6 13

E)

13 6

Algebra 10

9. Which expression is equivalent to

13. In the figure, m(∠ABC) = 90°, AD = 2, DB = 4 and BC = 3. Find sin θ.

⎛ 3π ⎞ ⎛ 17 π ⎞ cos ⎜ + x ⎟ + sin ⎜ – x ⎟+ tan (2 π – x)? ⎝ 2 ⎠ ⎝ 2 ⎠

A) cos x + sin x + tan x B) cos x – sin x – tan x

A 2 D

C) – cos x + sin x – tan x

4

q

D) – cos x – sin x + tan x E) cos x + sin x – tan x

B

A)

2

B)

5 5

11

C)

5 5

C

3

2 11

D)

11 2

E)

2 5

10. Which one of the following is false? A) sin 40° = cos 50° B) tan 210° = cot 30°

π is given. 3 Evaluate (sin a – sin b)2 + (cos a + cos b)2.

14. a + b =

C) tan 45° = cot 225° D) sec 89° = csc 1° E) sin 280° = –cos 10°

11. Find the value of A) –ñ3

B) −

A) 0

sin 15 ° cos 15 ° + . sin 45 ° cos 45 °

3 2

C) 0

D)

3 2

E) ñ3

B) 1

C) 2

15. In the figure, ABCD is a square and 3 ⋅ DE = AB. Find csc α.

D) 3

E) 4

D

E

C

F a B

A

A) ⎡π ⎣2

12. x ∈ [0, π], y ∈ ⎢ ,

1 5 2

16. cos x =

− 3 −2 2 6

D) Chapter Review Test 3B

B) 1 6

5 2 7

C) 5ñ2

D)

7

E) 7

5 2

3π ⎤ 1 1 , cos x = – and sin y= 2 ⎥⎦ 3 2

are given. Find cos (x + y). A)

B)

3+2 2 6

E)

C)

2 2− 3 6

3−2 2 6

sin 2x. A)

1 is given. If x is an acute angle, find 3

2 2 2

B)

2 2 3

C)

2 9

D)

4 2 9

E)

8 2 9

265

3C

CHAPTER REVIEW TEST 1. Find the primary directed angle of − A)

2π 5

B)

3π 5

4π 5

C)

D)

5. In the figure, ABCD is

72 π . 5

7π 5

E)

8π 5

⎛ 1 ⎛ ⎞ ⎛ 3⎞ A) ⎜ − , B) ⎜ − 2 , 2 ⎟ C) ⎜ 2 , − 2 ⎜ 2 2 ⎟⎟ ⎜ 2 ⎟ ⎜ 2 2 ⎠ 2 ⎝ ⎠ ⎝ ⎝ ⎛ ⎞ D) ⎜ 1 , − 3 ⎟ ⎜2 2 ⎟⎠ ⎝

⎛ ⎞ E) ⎜ − 2 , − 2 ⎟ ⎜ 2 2 ⎟⎠ ⎝

3. In the figure, ABC

A

3 4

B)

1

A)

10

B)

3 10

B

1 3

C)

6. In the figure, ABC is a

D)

1 6

E)

10 3

A

right triangle with m(∠B) = 90°, m(∠BAC) = 45°, m(∠ACD) = 15° and AC = 2ñ6. Find DC.

A

is a right triangle with altitude AH. CH = 4 and AC = 6 are given. Find sin B. A)

⎞ ⎟⎟ ⎠

C

a

rectangle with AB = 3 ⋅ AD. What is a possible value of tan α?

2. Find the coordinates of the terminal point of an 3π arc in standard position with length on the 4 unit circle.

D

45° D

2ñ6

15°

6

C

B B

2 3

H

1 2

C)

4. The figure contains five identical squares. What is sin α + cos β ?

C

4

D)

1 3

E)

A) 4

C) ñ6

B) 2ñ3

J

I

K

B

b

E) 5

1 4

7. 0 < x < π and cos x = – A

D) 2

3 are given. What is 5

csc x – sec x + tan x?

H

G

A)

19 12

B)

17 12

C)

7 12

D) −

13 6

E) −

7 6

a C

A)

2+ 5 10

D) 266

B)

1+ 2 10

D

1+ 5 10

E)

E

C) 2+ 3 10

F

3+ 5 10

8. cos 25° = a is given. Find sin 205° in terms of a. A) –a D) − 1 − a2

B) −

1 a

C)

1 − a2

E) a Algebra 10

9. cos 36° ≅ 0.8090 and cos 37° ≅ 0.7986 are given. Find cos (36° 15′). A) 0.8116

B) 0.8074

D) 0.8044

C) 0.8064 E) 0.8038

B) 0.0274

D) –0.0274

11 . cos α =

C) –0.0174 E) –0.0374

4 is given. If α is an acute angle, what is 5

1

B) −

10

4 5

C)

3 5

D)

4 10

A) tan 3a

B) cot 3a

E)

C) cot 4a E) tan 8a

14. 6x = π is given. Which one of the following is equivalent to A) 1

tan 5 x + tan x ? tan 5 x – tan x

B) –1

C) 0

D) 3x

E) 2x

1

π is given. Which one of the following is 20 sin 2 a – sin 4 a equivalent to ? sin 7 a ⋅ cos 9 a

10

A) –3

α sin ? 2 A) −

sin a + sin 3 a+ sin 5 a ? cos a + cos 3 a+ cos 5 a

D) tan 6a

10 . sin 36° ≅ 0.5878 and tan 42° ≅ 0.9004 are given. Find (3 ⋅ cos 54°) – (2 ⋅ cot 48°). A) 0.0374

13. Which one of the following is equivalent to

15. a =

B) –2

C) –1

D) 2

E) 3

12. sin 25° = a is given. Find the value of sin 40° in terms of a. A) 2a

2

B) 2a – 1 D)

1 – 2 a2 a2 +1

Chapter Review Test 3C

1 1 + ? cos 195 ° sin 195 °

16. What is 2

C) 1 – 2a E) 2a2 + 1

2

A) –2ñ2 D) –4ñ6

B) –4ñ2

C) –2ñ6 E) –6ñ2 267

CHAPTER REVIEW TEST

3D

1. What is 240° in radians? 3π A) 4

5π B) 3

5. A triangle ABC has side lengths a = 5 and b = 3.

7π C) 6

4π D) 3

sin x ⋅ tan x . cot 2 x

1 3

2. In a right triangle, cos x = . Find A)

64 3

D)

3. If A)

6 2 5

B) 17 3

11π E) 6

C)

81 2 2

8 2 3

E)

B) –

7 4

C)

7 4

D)

9 5

268

C)

21 2

E) ò17

D) 4

6. In a triangle ABC, m(∠A) = 60°, m(∠C) = 15°, and N is a point on BC such that AN bisects ∠A and BN = 4 cm. Find the length of AB. B) 3ñ3

C) 6ñ2

D) 2ñ3

E) –

8 5

A) 30°

B) 45°

C) 60°

D) 90°

E) 120°

8. x and y are acute angles. If sin x = B) cos x E)

C) 1

cos x sin x

E) 4

sin2A + sin2B = sin2C. Find m(∠C).

3 cos y = , find sin(x – y). 5

2

D) tan x

B) ò19

7. A triangle ABC has the property

sin x cos x ⋅ tan x cot x . 4. Simplify 1 − sin 2 x

A) 1 – tan x

A) 6

A) 4ñ2

3cos x + 2 sin x 2 = , what is cot x? 2 cos x − sin x 3 8 5

Find c if m(∠C) = 60°.

A)

33 65

B)

63 65

C) –

33 65

D)

13 48

5 and 13

E) –

14 65

Algebra 10

9. Find cos 75°.

13. What ratio is not equal to sin 15°?

6+ 2 4

A)

3− 2 2

B) 6 −2 4

D)

C)

A) sin 165° D) sin 105°

3 are given. 5

14. Simplify

Find sin 2x + cos 2x. 7 A) 25

17 B) – 25

11. Given that 0 < x < cos 2x. A)

A)

2 2

π 2 and sin x − cos x = , find 2 3

2 14 9

B)

12. Calculate 2 3

2 E) – 5

15. Simplify

D)

A)

27 D) – 25

7 C) 5

5 9

3 14 14

C) E)

13 13

3 4

Chapter Review Test 3D

C)

C) cos 75°

E) –cos 105°

3+ 2 4

A)

3 3

sin 410 ° ⋅ sin 210 ° ⋅ tan105 ° . cot( −15 °) ⋅ cos 220 °

B) 0

C)

1 2

D) –

1 2

E) 1

D) –

3 3

E) 1

sin105 ° + sin15 ° . cos105 °+ cos15 °

B) ñ3

C) –ñ3

4 9

sin15° ⋅ cos15 ° . tan15° ⋅ cot15 °

B)

B) –cos 255°

6− 2 4

E)

10. 90° < x < 180° and sin x =

2− 6 2

16. Simplify cos 20° ⋅ cos 40° ⋅ cos 60° ⋅ cos 80°. D)

1 4

E)

1 2

A)

1 16

B)

1 8

C)

3 8

D)

2 32

E)

269

1 2

CHAPTER REVIEW TEST 1. Calculate A)

1 2

3E 5. Write the signs of the trigonometric ratios

(sin0° ⋅ cos 90 °)+ tan0 ° − cos180 ° . 2sin 270 °+cot 90 °

B) –

1 2

C) 1

D) –1

sin 233°, cos 125°, tan 500° and sec 200° in order. E) 0

A) –, –, –, –

B) –, –, +, +

C) –, +, –, +

D) +, –, –, + E) –, +, –, –

2. Find the range of f(x) = 3sin 4x + 5. A) [3, 4]

B) [–1, 1]

D) [2, 8]

C) [–7, 17] E) [3, 5]

6. Which ratio is the biggest? A) sin 250°

B) cos 300°

D) cot 250°

C) sec 80°

E) tan 80°

π 3

3. Find the domain of f ( x) = 2 tan(2 x + ) − 5. π kπ + , k∈ 6 2 π kπ + , k∈ C) x ≠ 12 2

A) x ≠

E) x ≠

B) x ≠

π + kπ, k ∈ 6

D) x ≠

π + kπ, k ∈ 12

π + kπ, k ∈ 2

7. Find the fundamental period of f(x) = 3sin3 5x. A)

π 5

B)

π 3

C)

2π 3

D)

2π 15

E)

2π 5

4. In which expression is x undefined? A) cos x =

7 3

B) tan x = 28

C) sec x = 12

D) sin x =

E) cot x = 0.001 270

3 2

8. Find the fundamental period of f(x) = tan 3x + sin2 2x. A) 2π

B)

π 2

C)

5π 6

D) π

E)

π 6

Algebra 10

9. Find the fundamental period of f ( x) = sin A) π

B) π2

C) 2

D) 4

2x . π

E) 2π

12. Find the domain of f(x) = arccos(2x – 3). A) [1, 2]

B) [–1, 1]

C) [–5, –1]

E) (–∞, ∞)

D) [1, 5]

10. The graph of the function f(x) = sin x is given. Which of the following operations should be combined to draw the graph of f(x) = sin(2x – 3) + 1? I. move the graph 3 units left

1 2

13. Calculate arccos( − ). A)

II. move the graph 3 units right

π 3

B) –

π 3

C) –

5π 3

2π 3

D)

E)

π 6

III. move the graph 3 units down IV. move the graph 1 unit up V. move the graph 1 unit down VI. divide all values on the x–axis by 2

2 3

14. Calculate sin(arccos ).

VII. multiply all values on the x–axis by 2 A) I, IV, VII

B) I, V, VI

D) II, IV, VII

C) IV, II, VI

E) II, V, VI

11. What is the equation of the graph? y

p 2

A)

p

3p 2

–1

2 3

B)

1 3

C)

5 2

π 2

3 4

3 5

C) –

D) –

2 3

5 3

E)

15. Calculate cos( + arctan ). 2p

x

A)

4 5

B) –

4 5

D)

4 3

E)

3 5

–3 –5

A) y = 2cos 2x – 3

B) y = 2 sin 2x – 3

C) y = 3 cos 2x – 2

D) y = 3 sin 2x + 2

E) y = 3 cos 2x + 2

Chapter Review Test 3E

16. Calculate sin(arccos A)

33 65

B)

37 39

5 4 + arccot ). 13 3

C)

63 65

D)

13 65

E)

271

20 39

CHAPTER REVIEW TEST

3F

1. Solve 2cos 3x – 1 = 0.

6. Solve 3sin x – 5 cos x = 0. π 2kπ , k∈ B) ± + 9 3

π A) ± + 2kπ, k ∈ 9 π C) ± + 2kπ, k ∈ 3

π 2kπ D) ± + , k∈ 3 3

π 2 kπ + , k∈ 18 3

E) ±

5 A) arctan + kπ, k ∈ 3 4 B) arc cot + kπ, k ∈ 3 3 C) arcsin + 2 kπ, k ∈ 5 3 D) arctan + 2 kπ, k ∈ 5

2. Which of the following is not a solution of 2

2sin x – 1 = 0? A) 45°

B) 135°

C) 225°

D) 425°

E) 585°

3. What is the sum of the roots of 2cos2x – cos x = 0 in [0, 2π]? A) 2π

B)

5 E) arc cot + 2 kπ, k ∈ 3

7. What are the maximum and minimum values of f(x) = 4 sin x – 5 cos x?

11π 3

C) 4π

D)

17 π 3

E) 6π

A) 3, –3

B) 5, –5 D) 9, –9

π 3

C) ò41, –ò41

E) ñ7, –ñ7

π 4

4. Solve tan(3x − ) = tan . 7π kπ + , k∈ 12 2 π kπ + , k∈ C) 36 3

A)

E)

B)

π kπ + , k∈ 12 3

D)

7π kπ + , k∈ 36 3

3π kπ + , k∈ 4 3

8. Find the roots of sin 2x + cos x = 0 in [–π, π]. π 5π π π A) − , – , − , 2 6 6 2

C) −π, −

5. Solve sin x – cos x = 1. A)

π + 2 kπ, k ∈ 2

C) 2kπ, k ∈

π ,π 6

π π π π D) − , − , − , 2 3 6 2 π π π π E) − , − , , 4 6 4 2

π B) { + kπ, 2 nπ}, k, n ∈ 2 π D) { + 2 kπ, 2 nπ}, k, n ∈ 2

π E) { + 2kπ, π + 2 nπ}, k, n ∈ 2 272

π π π π B) − , − , , 2 6 6 2

2 9. Find the sum of the roots of cot x − 3 = 0 in

[0°, 540°].

tan x +1

A) 1130° B) 720° C) 630° D) 1620° E) 1480° Algebra 10

⎧sin x + cos y =1 π. x+ y = ⎪⎩ 2 π π A) ( + 2 kπ, − 2 kπ), k ∈ 6 3 5π π π π B) ( + 2 kπ, − 2 kπ) ∪( +2 nπ, − − 2 nπ), k, n ∈ 6 3 6 3

10. Solve the system ⎪⎨

C) (

π 5π + 2kπ, − − 2kπ), k ∈ 6 3

π D) ( + 2 kπ, 2 kπ), k ∈ 2 π π π E) ( + 2 kπ, 2 kπ) ∪ ( + 2 n π, + 2 n π), k, n ∈ 2 6 3

11. Solve sin x ≤

2 . 2

π ⎡ π ⎤ A) ⎢ − + 2kπ, + 2kπ ⎥ , k ∈ 4 ⎣ 4 ⎦ 3π ⎡π ⎤ + 2 kπ ⎥ , k ∈ B) ⎢ + 2kπ, 4 4 ⎣ ⎦ π ⎡ 5π ⎤ C) ⎢ − + 2kπ, + 2k π ⎥ , k ∈ 4 ⎣ 4 ⎦ 3π ⎡ 5π ⎤ D) ⎢ − + 2kπ, − + 2k π ⎥ , k ∈ 4 4 ⎣ ⎦ 3π ⎡π ⎤ E) ⎢ + 2kπ, + 2 kπ ⎥ , k ∈ 4 ⎣4 ⎦

π 3

12. Solve cos(3 x − ) ≥

3 . 2

π ⎡π ⎤ A) ⎢ + 2kπ, + 2kπ ⎥ , k ∈ 6 ⎣18 ⎦ π ⎡π ⎤ B) ⎢ + 2 kπ, + 2 kπ ⎥ , k ∈ 2 ⎣6 ⎦ 2 2 π π π π k k ⎡ ⎤ , + , k∈ C) ⎢ + 3 6 3 ⎥⎦ ⎣9 ⎡ π 2kπ π 2kπ ⎤ D) ⎢ − + , + , k∈ 3 6 3 ⎥⎦ ⎣ 18 ⎡ π 2kπ π 2kπ ⎤ E) ⎢ + , + , k∈ 3 6 3 ⎥⎦ ⎣18 Chapter Review Test 3F

π 6

13. Solve cot(2 x + ) − 3 < 0 . 5π ⎛ ⎞ A) ⎜ kπ, + kπ ⎟ , k ∈ 6 ⎝ ⎠ ⎛ kπ 5π kπ ⎞ + B) ⎜ , ⎟, k ∈ 2 ⎠ ⎝ 2 12 5π ⎛π ⎞ C) ⎜ + kπ, + kπ ⎟ , k ∈ 6 6 ⎝ ⎠ ⎛π ⎞ D) ⎜ + kπ, π + kπ ⎟, k ∈ ⎝6 ⎠ ⎛ π kπ 5π kπ ⎞ E) ⎜ + , + ⎟, k ∈ 2 ⎠ ⎝ 12 2 12

14. Solve –1 < tan x ≤ ñ3. 3π ⎛π ⎤ A) ⎜ + kπ, + kπ ⎥ , k ∈ 4 ⎝3 ⎦ π ⎛ π ⎤ B) ⎜ − + kπ, + kπ ⎥ , k ∈ 6 ⎦ ⎝ 4 π ⎛ π ⎤ C) ⎜ − + kπ, + kπ ⎥ , k ∈ 3 ⎦ ⎝ 4 π ⎞ ⎡ π D) ⎢ − + kπ, + kπ ⎟ , k ∈ 4 ⎣ 3 ⎠ π ⎛ π ⎤ E) ⎜ − + 2kπ, + 2kπ ⎥ , k ∈ 3 ⎦ ⎝ 4

15. Solve 2cos2x + 3cos x – 2 > 0. π ⎛ π ⎞ A) ⎜ − + 2kπ, + 2 kπ ⎟, k ∈ 3 3 ⎝ ⎠ π ⎛ π ⎞ B) ⎜ − + 2kπ, + 2 kπ ⎟, k ∈ 6 6 ⎝ ⎠ 2π ⎛π ⎞ C) ⎜ + 2 kπ, + 2 kπ ⎟, k ∈ 3 3 ⎝ ⎠

D) ( − arccos 2+ 2 kπ,arccos 2+ 2 k π), k ∈ π ⎛ π ⎞ E) ⎜ − + kπ, + 2 kπ ⎟ , k ∈ 3 ⎝ 2 ⎠ 273

APPLICATÝON SAYFASI GELECEK

274

Algebra 10

A. TANGENTS The word ‘tangent’ comes from the Latin word tangens, which means ‘touching’. Thus, a tangent line to a curve is a line that “just touches” the curve. In other words, a tangent line should be parallel to the curve at the point of contact. How can we explain this idea clearly? Look at the figures below.

y

A

a tangent line to a curve

A

zoomed in once

zoomed in twice

B original curve

y

A

A

x

B

As we zoom in to the curve near the point A, the curve becomes almost indistinguishable from the tangent line. So, the tangent line is parallel to the curve at the point A.

A

x a secant line to a curve

How can we find the equation of a tangent to a curve at a given point? The graphs below show one approach. y

t

y B(x, f(x))

B B

f(x) – f(a) A(a, f(a)) The slope of the line is the tangent of the angle between the line and the positive x-axis.

a positive slope

zero slope

276

a negative slope

no slope

A

B

x–a

a

x

x

x

The first graph shows the curve y = f(x). The points A(a, f(a)) and B(x, f(x)) are two points on this curve. The secant line AB has slope mAB, where mAB =

f (x) − f (a) . x−a

Now suppose that we want to find the slope of the tangent to the curve at point A. The second graph above shows what happens when we move point B closer and closer to point A on the curve. We can see that the slope of the secant line AB gets closer and closer to the slope of the tangent at A (line t). In other words, if m is the slope of the tangent line, then as B approches A, mAB approaches m.

Algebra 10

Definition

tangent line The tangent line to the curve y = f(x) at the point A(a, f(a)) is the line through A with the slope m = lim x→ a

f (x) − f (a ) , x−a

provided that this limit exists. Example

1

Solution

Find the equation of the tangent line to the curve y = x2 at the point A(1, 1). We can begin by calculating the slope of the tangent. Here we have a = 1 and f(x) = x2, so the slope is m = lim

f (x) − f (1) x2 − 1 = lim x →1 x − 1 x −1

m = lim

(x − 1)(x +1) = lim( x+1) =1+1= 2. x →1 (x − 1)

x →1

The equation of a line through the point (x1, y1) with slope m: y – y1 = m(x – x1).

x →1

Now we can write the equation of the tangent at point (1, 1): y – y1 = m(x – x1) y – 1 = 2(x – 1) y = 2x – 1.

Example

2

Solution

Find the equation of the tangent line to the curve y = x3 – 1 at the point (–1, –2). Here we have a = –1 and f(x) = x3 – 1, so the slope is m = lim

x→ –1

x3+ y3=(x+ y)(x2–xy+y2)

f ( x) – f ( −1) ( x3 − 1) − (( −1)3 − 1) x3 +1 = lim = lim x→ –1 x→ –1 x +1 x – (–1) x +1

( x +1)( x2 – x + 1) x→ –1 ( x +1)

m = lim

m = lim( x2 – x +1) = (–1) 2 – (–1)+1 x→ –1

m = 3.

So the equation of the tangent line at (–1, –2) with slope m = 3 is y – y1 = m(x – x1) y – (–2) = 3(x – (–1)) y + 2 = 3x + 3 y = 3x + 1. Differentiation

277

We can also write the expression for the slope of a tangent line in a different way. Look at the graphs below. y

y B

B

y=f(x)

A

a

y=f(x)

A a+h

h h h h

x

x

From the first graph, writing x = a + h gives us the slope of the secant line mAB =

f (a+ h) − f (a) . h

We can see in the second graph that as x approaches a, h approaches zero. So the expression for the slope of the tangent line becomes: f (a + h) − f (a ) m = lim . h→0 h THE SLOPE OF A TANGENT LINE TO A CURVE The slope of a tangent line to a curve y = f(x) at x = a is m = lim h→0

Example

3

Solution

Find the equation of the tangent line to the curve y = x3 at the point (–1, –1). Let f(x) = x3. Then the slope of the tangent at (–1, –1) is m = lim h→0

(x+y)3=x3+3x2y+3xy2+y3

f (a + h) − f (a ) . h

f ( − 1+ h) − f (–1) ( −1+ h) 3 −( −1) 3 = lim h→0 h h

(–1)3 + 3(–1) 2 h+ 3(–1) h2 + h3 −(–1) 3 h→0 h

m = lim m = lim h→0

h(3 − 3h + h2 ) = lim(3 − 3 h+ h2 ) = 3. h→0 h

So, the equation of the tangent at point (–1, –1) is y – (–1) = 3(x – (–1)) y + 1 = 3x + 3 y = 3x + 2. 278

Algebra 10

Example

4

Solution

Find the equation of the normal line to the curve y =

2 at the point (2, 1). x

Recall that a normal line is a line which is perpendicular to a tangent. The product of the slopes mt of a tangent and mn of a normal is –1. Let us begin by finding the slope of the tangent. 2 2 2 – –1 f (2+ h) – f (2) mt = lim = lim 2+ h 2 = lim 2+ h h→ 0 h→ 0 h→ 0 h h h mt = lim h→ 0

2 – (2+ h) –1 –h = lim = lim h → 0 h → 0 h( h + 2) h( h + 2) 2+ h

1 mt = – . 2

We have mt ⋅ mn = –1. mn ⋅ mt = –1 The product of slopes of the tangent line and the normal line at a point equals –1.

So, mn =

–1 –1 = = 2. 1 mt – 2

The equation of the normal line passing through the point (2, 1) with the slope mn = 2 is y – y1 = mn(x – x1) y – 1 = 2(x – 2) y = 2x – 3.

Check Yourself 1 1. Find the equation of the tangent line to each curve at the given point P. a. f(x) = x2 – 1

P(–1, 0)

b. f(x) = x3 + 1 P(0, 1) 1 1 P( , 2) c. f(x) = 2 x 2. Find the equation of the normal line at point P for each curve in the previous question. Answers 1. a. y = –2x – 2 b. y = 1 c. y = –4x + 4 2. a. y = Differentiation

1 1 x+ 2 2

b. x = 0 c. y = 1 x + 15 4 8 279

B. VELOCITIES

position at time t = 2

Imagine you are in a car driving across a city. The velocity of the car will not be constant. Sometimes the car will travel faster, and sometimes it will travel slower. However, the car has a definite velocity at each moment. This is called the instantaneous velocity of the car. How can we calculate the instantaneous velocity?

position at time t = 2+h

x

0

S(2 + h) – S(2) S(2) S(2+h)

To answer this question, let us look at a simpler example: the motion of an object falling through the air. Let g = 9.8 m/s2 be the acceleration of the object due to gravity. We know from physics that after t seconds, the distance that the object will have fallen is 1 2 gt meters or s(t) = 4.9 t2 meters. 2 Suppose we wish to calculate the velocity of the object after two seconds. We can begin by

s(t) =

calculating the average velocity over the time interval [2, 2 + h]: average velocity =

=

distance travelled elapsed time s(2 + h) − s(2) 4.9(4 + 4h+ h2 − 4) = = 19.6 + 4 .9 h h h

If we shorten the time period, the average velocity is becoming closer to 19.6 m/s, the value of instantaneous velocity. More generally, we can calculate the instantaneous velocity V(a) of an object at time t = a by the limit of the average velocities: V(a ) = lim h→ 0

s(a + h) − s( a) h

This is not the first time we see the above formula. It is the same formula that we use for the slope of the tangent line to a curve. Remember that f (a + h) − f (a ) . h This means that the velocity at time t = a is equal to the slope m = lim h→ 0

of the tangent line at A(a, s(a)). 280

Algebra 10

Example

5

Solution

A stone is dropped from the top of the Eiffel Tower. What is the velocity of the stone after five seconds? We use the equation of motion s(t) = 4.9t2 to find the velocity V after five seconds: V(5) = lim

s(5+ h) − s(5) 4.9(5+ h) 2 − 4.9(5) 2 = lim h→0 h h

V(5) = lim

4.9(25 +10 h+ h2 − 25) 4.9(10 h+ h2) = lim h→0 h h

V(5) = lim

4.9h(10+ h) = lim(49+ 4.9 h) = 49 m/s. h→0 h

h →0

h →0

h →0

Example

6

A particle moves along a straight line with the equation of motion s(t) = t2 + 3t + 1, where s(t) is measured in meters and t is in seconds. a. Find the average velocity over the interval [1, 2]. b. Find the instantaneous velocity at t = 2.

Solution

a. Average velocity is the ratio of distance travelled to elapsed time. So, we have average velocity =

s(2) – s(1) (2 2 + 3 ⋅ 2+1) – (1 2 + 3 ⋅1+1) = = 6 m/s. 1 2 –1

b. Let V(2) be the velocity after two seconds. V(2) = lim

s(2+ h) – s(2) ( h+ 2 )2 + 3 ⋅( h + 2) +1 −[2 2 + 3 ⋅2 +1] = lim h→ 0 h h

V(2) = lim

h2 + 4h + 4+ 3h +6+1 − 11 h2 +7 h = lim = lim( h+7) = 7 m/s. h→ 0 h→ 0 h h

h→ 0

h→ 0

Check Yourself 2 1. A basketball player throws a ball upward at a speed of 20 m/s. This means that after t seconds, the ball’s height will be s(t) = 20t – 4.9t2. a. Find the average velocity of the ball over the interval [1, 2]. b. Find the instantaneous velocity of the ball after two seconds. 2. The displacement of a particle moving in a straight line is given by the equation of motion s(t) = 2t3 + 3t – 2, where t is measured in seconds and s(t) is in meters. a. Find the average velocity of the particle over the following intervals. i. [1, 3] ii. [1, 4] iii. [2, 4] b. Find the instantaneous velocity of the particle at each time. i. t = 2 ii. t = 3 iii. t = 4 Answers 1. a. 5.3 m/s b. 0.4 m/s 2. a. i. 29 m/s ii. 45 m/s iii. 59 m/s b. i. 27 m/s ii. 57 m/s iii. 99 m/s Differentiation

281

C. RATES OF CHANGE In section A we learned how to find the slope of a tangent line and in section B we learned how to calculate the instantaneous velocity of an object from a given acceleration. We can say that acceleration is a rate of change: it shows how fast or slowly a quantity (the velocity) changes from one moment to the next. Other examples of rates of change are how fast a population grows, or how fast the temperature of a room changes over time. The problem of finding a rate of change is mathematically equivalent to finding the slope of a tangent line to a curve. To understand why, suppose y is a quantity that depends on another quantity x. Thus, y is the function of x and we write y = f(x). Look at the graph of f(x). If x increases by an amount h, then y increases by f(x + h) – f(x). y

f(x+h)

y=f(x) B(x+h, f(x+h)) f(x+h) – f(x)

f(x)

A(x, f(x))

f(x+h) – f(x) is the change in y that corresponds to a change h in x.

h x

The difference quotient

x+h

x

f (x+ h) − f (x) is called the average rate of change of y with h

respect to x over the interval [x, x + h] and can be interpreted as the slope of the secant line AB. If we take the limit of the average rate of change, then we obtain the instantaneous rate of change of y with respect to x, which is interpreted as the slope of the tangent line to the curve y = f(x) at A(x, f(x)). The following summarizes this part: RATES OF CHANGE 1. The average rate of change of f over an interval [x, x + h] is f (x+ h) − f (x) . h 2. The instantaneous rate of change of f(x) at a point x is f (x + h) − f (x) lim . h→0 h 282

Algebra 10

Example

7

Solution

A student begins measuring the air temperature in a room at eight o’clock in the morning. 2 She finds that the temperature is given by the function f (t ) =16+ t 2 °C, where t is in 3 hours. How fast was the temperature rising at 11:00? We are being asked to find the instantenous rate of change of the temperature at t = 3, so we need to find the following limit: f (3+ h) – f (3) h 2 2 16 + (3 + h)2 – (16 + (3) 2 ) 3 3 = lim h →0 h 2 2 16+ ⋅ (9+6 h + h2 ) – 16 − ⋅ 9 3 3 = lim h →0 h 2 (9 + 6 h + h2 – 9) 3 = lim h →0 h 2h(6+ h) = lim h →0 3h 2 = lim(6+ h) 3 h →0 2 = ⋅ 6 = 4 °C per hour 3

rate of change = lim

h →0

Example

8

Solution

A manufacturer estimates that when he produces x units of a certain commodity, he earns R(x) = x2 – 3x – 1 thousand dollars. At what rate is the revenue changing when the manufacturer produces 3 units? We need to find the instantaneous rate of change of the revenue at x = 3, so rate of change = lim

h→ 0

R(3+ h) – R(3) (3+ h) 2 – 3(3+ h) – 1 – (3 2 – 3 ⋅3 –1) = h h

9+6 h + h2 – 9 – 3 h 3 h + h2 = lim = lim(3+ h) = 3. h→ 0 h→ 0 h→ 0 h h It follows that revenue is changing at the rate of $3000 per unit when 3 units are produced. = lim

In conclusion, rates of change can be interpreted as the slope of a tangent. Whenever we solve a problem involving tangent lines, we are not only solving a problem in geometry but also solving a great variety of problems in science. Differentiation

283

D. DERIVATIVE OF A FUNCTION f (x+ h) − f (x) as a ‘difference quotient’ of the h function f(x). We have calculated the limit of a difference quotient as h approaches zero.

Up to now we have treated the expression

Since this type of limit occurs so widely, it is given a special name and notation. derivative of a function

Definition

The derivative of the function f(x) with respect to x is the function f ′(x) (read as “f prime of x”) defined by f ′(x) = lim h→ 0

f (x + h) − f (x) . h

The process of calculating the derivative is called differentiation. We say that f(x) is differentiable at c if f ′(c) exists. Thus, the derivative of a function f(x) is the function f ′(x), which gives 1. the slope of the tangent line to the graph of f(x) at any point (x, f(x)), 2. the rate of change of f (x) at x. FOUR-STEP PROCESS FOR FINDING f′(x) 1. Compute f(x + h). 2. Form the difference f(x + h) – f(x). 3. Form the quotient

f (x+ h) − f (x) . h

4. Compute f ′(x) = lim h→ 0

Example

9

Solution

f (x + h) − f (x) . h

Find the derivative of the function f(x) = x2. To find f ′(x), we use the four-step process: 1. f(x + h) = (x + h)2 = x2 +2xh + h2 2. f(x + h) – f(x) = x2 + 2xh + h2 – x2 = 2xh + h2 2 3. f (x+ h) − f (x) = 2 xh+ h = h(2x+ h ) = 2x+ h h h h

4. lim(2 x + h) = 2 x h→ 0

Thus, f ′(x) = 2x. 284

Algebra 10

Example

10

Solution

Find the derivative of the function f(x) = x2 – 8x + 9 at x = 1. We apply the four-step process: 1. f(x + h) = (x + h)2 – 8(x + h) + 9 = x2 + 2xh + h2 – 8x – 8h + 9 2. f(x + h) – f(x) = x2 + 2xh + h2 – 8x – 8h + 9 – (x2 – 8x + 9) = 2xh + h2 – 8h 2 3. f (x+ h) − f (x) = h + 2 xh − 8h = h + 2x − 8 h h

4. lim h→0

f (x + h) − f (x) = lim( h + 2 x − 8) = 2 x − 8 h→0 h

So, f ′(x) = 2x – 8 and f ′(1) = 2 ⋅ 1 – 8 = –6. This result tells us that the slope of the tangent line to the graph of f(x) at the point x = 1 is –6. It also tells us that the function f(x) is changing at the rate of –6 units per unit change in x at x = 1.

Example

11

Let f(x) =

1 . x

a. Find f ′(x). b. Find the equation of the tangent line to the graph of f(x) at the point (1, 1).

Solution

h 1 1 − − f (x + h) − f (x) 1 1 x x ( + h) a. f ′(x) = lim = lim x + h x = lim = lim( − )= − 2 . 0 0 0 h→0 h → h → h → h h h x(x + h) x

b. In order to find the equation of a tangent line, we have to find its slope and one point on the tangent line. We know that the derivative gives us the slope of the tangent. Let m be the slope of the tangent line, then 1 = −1. 12 So, the equation of the tangent line to the graph m = f ′(1) = −

y = –x+2

y

2 y=

1

y=

1 x

1

2

1 x x

of f(x) at the point (1, 1) with the slope m = –1 is y – 1 = –1(x – 1) y = –x + 2. Differentiation

285

Example

12

Solution

The function f(x) = ñx is given. Find the derivative of f(x) and the equation of the normal line to f(x) at the point x = 1. ⎛ x + h – x x+ h + x ⎞ f (x + h) − f (x) x+ h − x = lim = lim ⎜ ⋅ ⎟ h→ 0 h→ 0 h h h x+ h + x ⎠ ⎝

f ′(x) = lim h→ 0

f ′(x)=lim h→ 0

f ′(x)=

x+ h − x h( x + h + x )

= lim h→ 0

h h( x + h + x )

= lim h→ 0

1 x+ h + x

1 2 x

Remember that if mt is the slope of a tangent and mn is the slope of a normal at the same point, then mt ⋅ mn = –1. So, we can find the slope of the normal from the slope of the tangent. Then we can write the equation of normal line to f(x) at the point x = 1. The slope of the tangent is 1 1 = . 2⋅ 1 2 The slope of the normal is mt = f ′(1) =

mn = −

1 1 = − = −2. 1 mt 2

The equation of the normal line is y – y0 = mn ⋅ (x – x0) y – 1 = –2(x – 1)

(Note that y0 = f(x0), that is y0 = f(1) = 1)

y = –2x + 3.

Check Yourself 3 1. Find the derivative of the function f(x) = 2x + 7. 2. Let f(x) = 2x2 – 3x. a. Find f ′(x). b. Find the equation of the tangent line to the graph of f(x) at the point x = 2. 3. Find the derivative of the function f(x) = x3 – x. 1 4. If f ( x) = , find the derivative of f(x). x+ 2 Answers 1. 2 2. a. 4x – 3 b. y = 5x – 8

286

3. 3x2 – 1

4. −

1 2 ( x + 2)3 Algebra 10

E. LEFT-HAND AND RIGHT-HAND DERIVATIVES When we were studying limits we learned that the limit of a function exists if and only if the left-hand and the right-hand limits exist and are equal. Otherwise the function has no limit. From this point, we may conclude that the derivative of a function f(x) exists if and only if f ( x + h) − f ( x ) f ( x + h) − f ( x ) and f ′( x+ ) = lim+ exist and are equal. h→ 0 h h hand derivative and the right-h hand These expressions are respectively called the left-h derivative of the function. f ′( x− ) = lim− h→ 0

Example

13

Solution

Show that the function f(x)= ñx does not have a derivative at the point x = 0. Here we should find the left-hand derivative and the right-hand derivative. If they exist, then we will check whether they are equal or not. Let us find the left-hand derivative: f ′(0 − ) = lim− h→ 0

f (0 + h) − f (0) 0 +h − 0 h = lim− = lim− . h→ 0 h→ 0 h h h

Since h < 0, ñh is undefined and this limit does not exist. So the left-hand derivative does not exist either. Thus, the function f(x)= ñx has no derivative at the point x = 0.

Example

14

Solution

⎧⎪ x2 − 1, x ≥1 . f(x) is given as f ( x) = ⎨ ⎪⎩2 x − 2, x < 1 Does this function have a derivative at the point x = 1?

We will find the left-hand and the right-hand derivatives. f ′(1− ) = lim−

f (1 + h) − f (1) 2(1 + h) − 2 − 0 2h = lim− = lim− =2 h→ 0 h→ 0 h h h

f ′(1+ ) = lim+

f (1 + h) − f (1) (1 + h) 2 − 1 − 0 h2 + 2 h = lim+ = lim+ = lim( h + 2 ) = 2. h→ 0 h→ 0 h→ 0 + h h h

h→ 0

h→ 0

The left-hand and the right-hand derivatives are equal to each other. Thus, the derivative of the function at the point x = 1 exists and f ′(1) = f ′(1–) = f ′(1+) = 2. Differentiation

287

F. DIFFERENTIABILITY AND CONTINUITY Recall that if f ′(c) exists, then the function f(x) is differentiable at point c. Similarly, if f(x) is differentiable on an open interval (a, b), then it is differentiable at every number in the interval (a, b). Example

15

Solution

Where is the function f(x) = |x| differentiable? We can approach this problem by testing the differentiability on three intervals: x > 0, x < 0 and x = 0. 1. If x > 0, then x + h > 0 and |x + h| = x + h. Therefore, for x > 0 we have f ′(x) = lim h→ 0

| x + h | – | x| x+ h – x h = lim = lim = lim1=1. h → 0 h → 0 h h h h→ 0

So, f ′(x) exists and f (x) is differentiable for any x > 0. 2. If x < 0, then |x| = –x and |x + h| = –(x + h) if we choose h small enough such that it is nearly equal to zero. Therefore, for x < 0 we have f ′(x) = lim x→0

| x + h | – | x| –( x + h) – (– x) –h = lim = lim = lim(–1) = –1 x → 0 h → 0 h→ 0 h h h

So, f ′(x) exists and f(x) is differentiable for any x < 0. 3. For x = 0 we have to investigate the left-hand and the right-hand derivatives separately: lim+

|0+ h| – |0| | h| h = lim+ = lim+ = lim+ 1=1 h→ 0 h→ 0 h h→ 0 h h

lim

|0+ h| – |0| | h| –h = lim– = lim– = lim(–1) = –1 h→ 0 h→ 0 h h→ 0– h h

h→ 0

h→ 0–

Since these limits are different, f ′(x) does not exist. So, f(x) is not differentiable for x = 0. In conclusion, f(x) is differentiable for all the values of x except 0. Alternatively, from the graph of f(x), we can see that f(x) does not have a tangent line at the point x = 0. So, the derivative does not

y

y = f(x)

exist. Note that the function does not have a derivative at the point where the graph has a ‘corner’. 288

x

Algebra 10

A function f is continuous at x = a if and only if lim f ( x) = f ( a).

If a function f(x) is differentiable at a point, then its graph has a non-vertical tangent line at this point. It means that the graph of the function cannot have a ‘hole’ or ‘gap’ at this point. Thus, the function must be continuous at this point where it is differentiable.

x→ a

Note If f(x) is differentiable at a, then f(x) is continuous at a. The converse, however, is not true: a continuous function may not be differentiable at every point. For example, the function f(x) = |x| is continuous at 0, because lim f (x) = 0 = f (0). x →0

But it is not differentiable at the point x = 0.

Example

16

⎧2 x2 − x, ⎪⎪ The piecewise function f(x) is given as f ( x) = ⎨6, ⎪ 3 ⎪⎩ x − 2, a. Is f(x) continuous at x = 2?

x>2 x = 2. x<2

b. Is f(x) differentiable at x = 2? Solution

a. Since lim f ( x) = f (2), f(x) is continuous at x = 2. x→ 2

b. Let us find the left-hand and the right-hand derivatives of the function f(x) at the point x = 2. f (2 − ) = lim− h→ 0

= lim− h→ 0

f (2+ h) − f (2) (2 + h)3 − 2 − 6 = lim− h→ 0 h h 2 3 + 3 ⋅ 2 2 h + 3 ⋅ 2 h2 + h3 – 8 h(12+6 h+ h2 ) = lim− h→ 0 h h

h+ h2 ) =12 = lim(12+6 − h→ 0

f (2 + ) = lim+ h→ 0

f (2+ h) − f (2) 2 ⋅(2+ h)2 − (2 + h) − 6 = lim+ h→ 0 h h

= lim+

8+8 h + 2 h2 − 2 − h − 6 2 ⋅ (4+ 4h + h2 ) − 2 − h − 6 = li m+ h→ 0 h h

= lim+

h(7+ 2 h) = lim(7+ 2 h) = 7. h→ 0 + h

h→ 0

h→ 0

Since f ′(2+) ≠ f ′(2–), the derivative of the function f(x) does not exist at the point x = 2. So, the function is continuous at x = 2, but it is not differentiable at the same point. Differentiation

289

We have seen that a function f(x) is not differentiable at a point if its graph is not continuous at x = a. The figures below show two more cases in which f(x) is not differentiable at x = a: y

y

a

y

x

a discontinuity

a

x

a corner

a

x

a vertical tangent

CRITERIA FOR DIFFERENTIABILITY For the following cases the function is not differentiable at a given point: 1. the graph has a discontinuity at the point, 2. the graph has a ‘corner’ at the point, 3. the graph has a vertical tangent line at the point.

Example

17

Solution

290

Explain why the function shown in the graph on the right is not differentiable at each of the points x = a, b, c, d, e, f, g.

y

y = f(x)

The function f(x) is not differentiable at the points x = a, b, c because it is discontinuous at each of these points. The derivative of the function f(x) does not exist at x = d, e, f d e f g x a b c because it has a corner at each of these points. Finally, the function is not differentiable at x = g because the tangent line is vertical at that point. Algebra 10

Check Yourself 4 1. Given that

⎧ x − 1, f ( x) = ⎨ 2 ⎩ x − 1,

x <1 , show that the derivative of f(x) does not exist at the x ≥1

point x = 1. 2. f(x) = |x2 – 4x + 3| is given. Find the derivative of f(x) at the point x = 3. 3. The graph of a function f is given below. State, with reasons, the values at which f is not differentiable. y y = f(x)

-1 1 2 3 4 5 6 7 8 9 10

11

x

Answers 1. compare f ′(1–) and f ′(1+). 2. does not exist. 3. x = –1, corner; x = 4, discontinuity; x = 8, corner; x = 11, vertical tangent.

Differentiation

291

At the beginning of our study of derivatives we have learned that a curve lies very close to its tangent line near the point of contact. This means that for the same value of x near the point of tangency, the values of y on the curve and tangent line are approximately equal to each other. This fact gives us a useful method for finding approximate values of functions. We can use the tangent line at (a, f(a)) as an approximation to the function f(x) when x is near a. The equation of this tangent line is y = f(a) + f ′(a)(x – a). So, our approximation becomes f(x) ≈ f(a) + f ′(a)(x – a). This type of approximation is called the linear approximation or tangent line approximation of f(x) at a. The linear function whose graph is the tangent line L(x) = f(a) + f ′(a)(x – a) is called the linearization of f(x) at a. The geometric interpretation of linear approximation is show in the figure. y = f(x)

y

y = f(x)

f(a) L(a)

a

x

The linear approximation f(x) ≈ L(x) is a good approximation when x is near a. It is very useful in physics for simplifying a calculation or a theory. You might think that a calculator can give us better approximation than the linear approximation. But a linear approximation gives an approximation over an entire interval, which can be more useful. For this reason, scientists frequently use linear approximation in their work. The following example illustrate the use of linear approximation method to simplify calculation.

For example, let us find the linearization of the function f ( x)= x + 2 at a = 2, and use it to approximate the numbers

4,01 .

3,99 and

First, we have to find f ′(2), the slope of the tangent line to the curve f ( x)= x + 2 when x = 2. 1 The derivative of f(x) is f ′( x)=( x + 2 )′ = . 2 x+ 2 1 1 So, f ′(2)= = . 2 2+2 4 The linearization is given by L(x) = f(a) + f ′(a) ⋅ (x – a)

1 x 3 (x – 2) = + . 4 4 2 x 3 The linear approximation is therefore x + 2 ≈ L( x)= + . 4 2 In particular we have

L(x) = f(2) + f ′(2) ⋅ (x – 2) = 2 +

3.99 = 1.99+ 2 ≈ L(1.99) = 4.01 ≈ L(2.01) =

1.99 3 + =1.9975 4 2

2.01 3 + = 2.0025. 4 2

x 3 The graphs of f ( x)= x + 2 and its linear approximation L( x)= + are shown below. We 4 2 see that our approximations are overestimates because the tangent line lies above the curve. y x 3 L(x) = + 4 2 2

–2

f(x) = x + 2

x

2

The following table shows estimates from the linear approximation with the actual values. approximation

actual value

3.99

1.9975

1.99749...

4.01

2.0025

1.00249...

EXERCISES

4 .1

A. Tangents 1. Find the slope of the tangent line to the graph of each function at the given point. a. f(x) = 5x – 1 ; x = 3 b. f(x) = 4 – 7x ; x = 2 c. f(x) = x2 – 1 ; x = –1 d. f(x) = 3x2 – 2x – 5 ; x = 0 › e. f(x) = x3 – 3x + 5 ; x = 1 › f.

f(x) = x + ñx ; x = 4

› g.

f ( x) =

› h. f ( x) =

1 ; x= 2 x2 4x ; x= 2 x +1

2. Find the equation of the tangent line to each

C. Rates of Change 5. The volume of a spherical cancer tumor is given 4 3 by the function V( r ) = π r , where r is the 3 radius of the tumor in centimeters. Find the rate

of change in the volume of the tumor when 2 r = cm. 3

6. A certain species of eagle faces extinction. After a conservation project begins, it is hoped that the eagle population will grow according to the rule N(t) = 2t2 + t + 100 (0 ≤ t ≤ 10), where N(t) denotes the population at the end of the year t. Find the rate of growth of the eagle population when t = 2 and the average rate of growth over the interval [2, 3].

function at the given point. a. f(x) = 2x + 5 at (2, 9) b. f(x) = x2 + x + 1 at (1, 3) › c. f(x) = x3 – x at (2, 6) › d. f(x) = 2ñx at (4, 4)

B. Velocities 3. A particle moves along a straight line with the equation of motion s(t) = t2 – 6t – 5, where s is measured in meters and t is in seconds. Find the velocity of the particle when t = 2.

4. If a stone is dropped from a height of 100 m, its

7. The fuel consumption (measured in litres per hour) of a car travelling at a speed of v kilometers per hour is c = f(v). a. What is the meaning of f ′(v)? b. What does the statement f ′(20) = – 0.05 mean?

height in meters after t seconds is given by s(t) = 100 – 5t2. Find the stone’s average velocity over the period [2, 4] and its instantaneous velocity at time t = 4. 294

Algebra 10

D. Derivative of a Function 8. Each limit below represents the derivative of a function f(x) at x = a. Find the function f and the number a in each case. (1+ h)10 − 1 h→ 0 h

3 b. lim 8+ h − 2 h→ 0 h cos(π + h)+1 d. lim h→0 h

a. lim

3x − 81 x→ 4 x − 4

c. lim

⎧ x +6, 14. Given that f ( x) = ⎪⎨x2 , ⎪ 3 ⎩ x – 6 x,

› c. f ( x) =

3

15. Consider the slopes of the tangent lines to the given curve at each of the five points shown. List these five slopes in decreasing order.

2 x +1 x −1

› d. f ( x) = 3x +1

x

›

0≤x≤8

6x,

x> 3

y A

E

B

D C x

16. If the tangent to the graph of f(x) = x2 – 2ax + 3

E. Left-Hand and Right-Hand Derivatives 10. Let f(x) =

x = 3 find f ′(3).

Mixed Problems

9. Find the derivative of each function. a. f(x) = 3 – 2x + x2 › b. f ( x) =

x> 3

. 9x – 24, 8 < x Does the function have derivative at x = 8? Why or why not?

17. At which point of the curve y = x2 + 4 does its

››

tangent line pass through the origin?

18. An arrow is shot upward on a planet. Its height (in ›

11. Given that f(x) = |x – 1|, find f ′(1).

at x = –1 is parallel to the line 2x – y = 1, find a.

meters) after t seconds is given by h(t) = 60t – 0.6t2.

F. Differentiability and Continuity

a. At what time will the arrow reach the top?

12.

b. With what velocity will the arrow hit the ground?

y

y = f(x)

–2 –1

1

3

6

19. Given the continuous function

››

x

find a, b and c such that its graph has a tangent touching it at three points..

The graph of f(x) is given. At what numbers is f(x) not differentiable? Why?

13. Let f(x) =

x2 + 7x, x ≤ 1

. 9x – 24, x > 1

Does the function have derivative at x = 1? Why or why not? Differentiation

⎧ x2 +10 x +8, x ≤ –2 ⎪ f ( x) = ⎨ax2 + bx + c, –2 < x < 0, ⎪ 2 x≥0 ⎩ x + 2 x,

20. Given that f(x) = |x2 – 2x|, find f ′(1). ›

21. Using linear approximation calculate ò99.

››

295

A. BASIC DIFFERENTIATION RULES Up to now, we have calculated the derivatives of functions by using the definition of the derivative as the limit of a difference quotient. This method works, but it is slow even for quite simple functions. Clearly we need a simpler, quicker method. In this section, we begin to develop methods that greatly simplify the process of differentiation. From now on, we will use the notation f′(x) (f prime of x) to mean the derivative of f with respect to x. Other books and mathematicians sometimes use different notation for the derivative, such as d dy = Dx( f (x)). f (x) = y′ = dx dx

All of these different types of notation have essentially the same meaning: the derivative of a function with respect to x. Finding this derivative is called differentiating the function with respect to x. In stating the following rules, we assume that the functions f and g are differentiable. Our first rule states that the derivative of a constant function is equal to zero. THE DERIVATIVE OF A CONSTANT FUNCTION If c is any real number, then c′ = 0. We can see this by considering the graph of the constant function f(x) = c, which is a horizontal line. The tangent line to a straight line at any point on the line coincides with the straight line itself. So, the slope of the tangent line is zero, and therefore the derivative is zero. We can also use the definition of the derivative to prove this result: f (x + h) − f (x) c −c f ′(x) = lim = lim = lim 0 = 0. h→ 0 h → 0 h→ 0 h h

Example

18

y y=c slope = 0

x The slope of the tangent to the graph of f(x) = c, where c is constant, is zero.

a. If f(x) = 13, then f ′(x) = (13)′ = 0. 1 ⎛ 1 ⎞′ b. If f (x) = − , then f ′( x) = ⎜ − ⎟ = 0 . 2 ⎝ 2⎠

296

Algebra 10

Next we consider how to find the derivative of any power function f(x) = xn. Note that the rule applies not only to functions like f(x) = x3, but also to those such as g(x) = 4 x3 and h( x) =

1 = x −5 . x5

THE DERIVATIVE OF A POWER FUNCTION (POWER RULE) If n is any real number, then (xn)′ = nxn – 1. Example

19

a. If f (x) = x, then f ′( x) = x ′ = 1 ⋅ x1−1 = 1 . b. If f (x) = x2 , then f ′( x) = ( x2 ) ′ = 2 ⋅ x 2 −1 = 2 x. c. If f (x) = x3 , then f ′( x) = ( x 3 ) ′ = 3 ⋅ x 3 −1 = 3 x 2 .

Note To differentiate a function containing a radical expression, we first convert the radical expression into exponential form, and then differentiate the exponential form using the Power Rule. Example

20

a. If f (x) = 2 x3 , then f ( x) = x 3 / 2 in exponential form f ′(x) = (x3 / 2 )′ =

3 3 / 2 −1 3 1 / 2 x = x . 2 2

1 b. If f (x) = , then f ( x) = x −1 in exponential form x f ′(x) = (x−1 )′ = −1 ⋅ x −1−1 = −x −2 = −

Differentiation

1 . x2

297

The proof of the Power Rule for the general case (n ∈ ) is not easy to prove and will no be given here. However, we can prove the Power Rule for the case where n is a positive integer.

Proof

f (x + h) − f (x) (x + h)n − x n = lim . h→ 0 h→ 0 h h Here we need to expand (x + h)n and we use the Binomial Theorem to do so:

(Power Rule) If f(x) = xn, then

f ′(x) = lim

n ⋅ (n − 1) n − 2 2 ⎡ n ⎤ n −1 x h + ⋅ ⋅ ⋅ + nxh n −1 + h n ⎥ − x n ⎢⎣ x + nx h + 2 ⎦ f ′(x) = lim h→ 0 h f ′(x) = lim

n ⋅ xn −1h +

h→ 0

n ⋅ (n − 1) n − 2 2 x h + ⋅ ⋅ ⋅+ nxh n −1 + h n 2 h

(every term includes h as a factor, so h’s can be simplified)

n ⋅ (n − 1) n − 2 ⎡ ⎤ f ′(x) = lim ⎢ nxn −1 + x h + ⋅ ⋅ ⋅+ nxh n − 2 + h n –1 ⎥ = n ⋅ x n −1 (if h = 0, then every term including h→ 0 h as a factor will be zero) 2 ⎣ ⎦

Check Yourself 5 Differentiate each function by using either the Constant Rule or the Power Rule. 1. f(x) = 2

2. f(x) = 0.5

5. f(x) = x3

3 6. f (x) = x7

3. f (x) = – 1 3

4. f (x) = 3 2

7. f (x) = 1 x2

8. f (x) =

1 x3

Answers 1. 0

2. 0

3. 0

4. 0

5. 3x2

6.

73 4 x 3

7. –

2 x3

8. −

3 2 x5

The next rule states that the derivative of a constant multiplied by a differentiable function is equal to the constant times the derivative of the function.

THE CONSTANT MULTIPLE RULE

[c ⋅

Example

21

f (x)]′= c ⋅ f ′(x)

,

c∈

a. If f (x) = 3x, then f ′( x) = (3 x) ′ = 3 ⋅ ( x) ′ = 3 ⋅ 1 = 3 . b. If f (x) = 3x4 , then f ′( x) = (3 x 4 ) ′ = 3( x 4 ) ′ = 3 ⋅ (4 x 3 ) =12 x 3 .

298

Algebra 10

Proof

(Constant Multiple Rule) If g(x) = c ⋅ f(x), then g′(x) = lim h→ 0

g(x + h) − g(x) c ⋅ f (x + h) − c ⋅ f (x) = lim h→ 0 h h

g′(x) = c ⋅ lim h→ 0

f (x + h) − f (x) h

g′(x) = c ⋅ f ′(x). Example

22

a. If f (x) = − b.

2 6 , then f ′( x) = (–2 x −3 ) ′ = −2( x −3) ′ = –2(–3 x −4) = 6 x −4 = 4 . 3 x x

5 ⎛1 ⎞ 5 . If f (x) = 5 x, then f ′( x) = (5 x1 / 2 ) ′ = 5( x1 / 2 ) ′ = 5 ⎜ x −1 / 2 ⎟ = x −1 / 2 = 2 x ⎝2 ⎠ 2

Next we consider the derivative of the sum or the difference of two differentiable functions. The derivative of the sum or the difference of two functions is equal to the sum or the difference of their derivatives. Note that the difference is also the sum since it deals with addition of a negative expression. THE SUM RULE

[ f (x) ∓ g(x)]′ = f ′(x) ∓ g′(x) Note We can generalize this rule for the sum of any finite number of differentiable functions.

[ f (x) ∓ g(x) ∓ h(x) ∓ ...]′ = f ′(x) ∓ g ′(x) ∓ h ′(x) ∓ ...

Differentiation

299

Now, let’s verify the rule for a sum of two functions. Proof

(Sum Rule) If S(x) = f(x) + g(x), then

[ f (x + h)+ g(x + h)] − [ f (x)+ g(x) ] S(x + h) − S(x) = lim h→ 0 h→ 0 h h

S′(x) = lim

S′(x) = lim

[ f (x + h) − f (x)]+ [g(x + h) − g(x) ] h

h→ 0

S′(x) = lim h→ 0

f (x + h) − f (x) g(x + h) − g(x) + lim h → 0 h h

S′(x) = f ′(x)+ g ′( x). Example

23

a. If f (x) = x−2 +7, then f ′( x) = ( x−2 +7) ′= ( x−2) ′+(7) ′= −2 x−3+0 = −2 x−3. 2 ′ ⎛ t2 2 b. If g(t) = t + 5 , then g′( t) = ⎛ t −2 ⎞ ⎜ +5t ⎟ = ⎜ 2 5 t ⎝5 ⎠ ⎝5

⎞′ 1 2 −2 −2 ⎟ +(5 t ) ′= ( t ) ′+5( t ) .′ 5 ⎠

1 g(t )′ = (2t 2–1 )+5(–2 t–2–1 ) 5 2 2t 10 g(t )′ = t − 10t −3 = − 3 . 5 5 t

Notice that in this example, the independent variable is t instead of x. So, we differentiate the function g(t) with respect to t. By combining the Power Rule, the Constant Multiple Rule and the Sum Rule we can differentiate any polynomial. Let us look at some examples.

Example

24

Solution

Differentiate the polynomial function f(x) = 3x5 + 4x4 – 7x2 + 3x + 6. f ′(x) = (3 x5 + 4 x4 − 7 x2 + 3 x +6)′ f ′(x) = (3 x5 )′+(4 x4 ) ′+(–7 x2 ) ′+(3 x) ′+(6) ′ f ′(x) = 3(x5 )′+ 4( x4 ) ′ − 7( x2 ) ′+ 3( x) ′+(6) ′ f ′(x) = 3 ⋅ 5 x4 + 4 ⋅ 4 x3 − 7 ⋅ 2 x+ 3 ⋅1+0 f ′(x) =15 x4 +16 x3 − 14 x+ 3

300

Algebra 10

Example

25

It is estimated that x months from now, the population of a certain community will be P(x) = x2 + 20x + 8000. a. At what rate will the population be changing with respect to time fifteen months from now? b. How much will the population actually change during the sixteenth month?

Solution

a. The rate of change of the population with respect to time is the derivative of the population function, i.e. rate of change = P′(x) = 2x + 20. Fifteen months from now the rate of change of the population will be: P′(15) = 2 ⋅ 15 + 20 = 50 people per month. b. The actual change in the population during the sixteenth month is the difference between the population at the end of sixteen months and the population at the end of fifteen months. Therefore, the change in population = P(16) – P(15) = 8576 – 8525 = 51 people.

Check Yourself 6 1. Find the derivative of each function with respect to the variable. 3 2x

b.

4 f (r ) = π r 3 3

d. f(x) = 3x2 + 5x – 1

e.

f (t ) =

a. f (x) =

2. Find the derivative of f (x) = 3. Differentiate f (x) = Answers 3 1. a. − 2 2x 2. 2x – 2 Differentiation

b. 4πr2

x x+ x 0.1 x

4 t2 − +t t3 3

3 2 f. f (x) = x − 4x + 3 x

x3 − 3x2 + 3x − 1 . x −1

x2 x − x

c.

c. f(x) = 0.2ñx

.

d. 6x + 5

e. –

12 2t – +1 3 t4

f. 2 x − 4 −

3 x2

3. 1 301

B. THE PRODUCT AND THE QUOTIENT RULES Now we learn how to differentiate a function formed by multiplication or division of functions. Based on your experience with the Constant Multiple and Sum Rules we learned in the preceding part, you may think that the derivative of the product of functions is the product of separate derivatives, but this guess is wrong. The correct formula was discovered by Leibniz and is called the Product Rule. The Product Rule states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function. THE PRODUCT RULE

[ f (x)g(x)]′ = f ′(x)g(x) + f (x)g ′(x) Be careful! The derivative of the product of two functions is not equal to the product of the derivatives: We can easily see this by looking at a particular example. Let f(x) = x and g(x) = x2. Then

[ f (x)g(x)]′ ≠ f ′(x)g′(x)

f (x)g(x) = x ⋅ x2 = x3

f ′(x) =1 and g′( x) = 2 x

[ f (x)g(x)] ′ = 3 x2

f ′(x)g′(x) =1 ⋅ 2 x = 2 x [ f (x)g( x)]′ ≠ f ′(x)g′(x).

Example

26

Solution

Find the derivative of the function f(x) = x(x + 1). By the Product Rule, f ′(x) = x ⋅ (x+ 1)′+ (x) ′ ⋅ (x+1) = x ⋅ 1+1 ⋅ ( x+1) = 2 x+1 .

We can check this result by using direct computation: f(x) = x(x + 1) = x2 + x so, f ′(x) = 2x + 1, which is the same result. Note that preferring direct differentiation when it is easy to expand the brackets is always simpler than applying the Product Rule. Example

27

Solution

Differentiate the function f(x) = (2x2 + 1)(x2 – x). f ′(x) = (2 x2 +1)′ ⋅ (x2 − x) +( 2x 2 + 1) ⋅ (x2 − x)′ f ′(x) = (4 x)(x2 − 1) + (2 x2 + 1)(2 x − 1) f ′(x) = 4x3 − 4x+ 4x3 − 2x2 + 2x − 1 f ′(x) = 8 x3 − 2 x2 − 2 x − 1

302

Algebra 10

28

Differentiate the function f(x) = (x3 + x –2)(2ñx + 1).

Solution

First, we convert the radical part into exponential form:

Example

f (x) = ( x3 + x – 2)(2 x +1) = ( x3 + x – 2) ⋅(2 x1/ 2 +1). Now, by the Product Rule, f ′(x) = (x3 + x – 2) ′ ⋅(2 x1/ 2 +1)+( x3 + x – 2) ⋅(2 x1/ 2 +1) ′ f ′(x) = (3 x2 +1)(2 x1/ 2 +1)+( x3 + x – 2) ⋅ x −1/ 2 = 6 x5 / 2 + 3 x2 + 2 x1/ 2 +1+ x5 / 2 + x1/ 2 – 2 x −1/ 2 f ′(x) = 7 x5 / 2 + 3x2 + 3x1 / 2 – 2 x–1/ 2 +1.

Let us look at the proof of the Product Rule. Proof

(Product Rule) If P(x) = f(x)g(x), then P′(x) = lim h→ 0

P(x + h) − P(x) f (x + h)g(x + h) − f (x)g(x) = lim h→ 0 h h

By adding –f(x + h)g(x) + f(x + h)g(x) (which is zero) to the numerator and factoring, we have: P′(x) = lim h→ 0

P′(x) = lim

f (x + h)g(x + h) − f (x + h)g(x ) + f (x + h)g(x ) − f (x)g(x) h f (x + h) [g(x + h) − g(x)]+ g(x) [f (x + h) − f (x) ] h

h→ 0

⎛ ⎡ g(x + h) − g(x) ⎤ ⎡ f (x + h) − f (x) ⎤ ⎞ P′(x) = lim ⎜ f (x + h) ⎢ + g( x) ⎢ ⎥ ⎥⎟ h→ 0 h h ⎣ ⎦ ⎣ ⎦⎠ ⎝ f (x + h) − f (x ) g(x + h) − g(x) + lim g(x) ⋅ lim h → 0 h → 0 h h ′ ′ ′ ′ ′ P (x) = f (x) ⋅ g (x)+ g(x) ⋅ f (x) = f (x)g(x)+ f (x)g (x). P′(x) = lim f (x + h) ⋅ l im h→ 0

Example

29

Solution

h→ 0

Differentiate the function f(x) = (x2 + 1)(3x4 – 5x)(x3 + 2x2 + 4). In this example we have a product of three functions, but we are only able to apply the rule for the product of two functions. So, before we proceed we must imagine the function as a product of two functions as follows: f ( x) = ( x2 +1)(3 x4 − 5 x) ( x3 + 2 x2 + 4) f ′( x) = [( x2 +1)(3 x4 − 5 x)]′ ( x3 + 2 x2 + 4) +( x2 +1)(3x 4 − 5 x)( x3 + 2 x2 + 4)′ requires product rule once more

f ′( x) = [2 x(3x4 − 5x)+( x2 +1)(12 x3 − 5)] ( x3 + 2 x2 + 4) +( x2 +1)(3 x4 − 5 x)(3x2 + 4x ).

Our aim is to introduce this method and because any further simplification is time consuming, we will stop at this point. Differentiation

303

The derivative of a quotient is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator. Or, ⎛ numerator ⎞′ derivative of the numerator ×denominator – numerator × derivative of the denominator . ⎜ ⎟ = the square of the denominator ⎝ denominator ⎠

THE QUOTIENT RULE ⎛ f (x) ⎞′ f ′(x)g(x) − f (x)g ′(x) ⎜ g(x) ⎟ = ( g(x))2 ⎝ ⎠

,

g( x) ≠ 0

The quotient rule is probably the most complicated formula you will have to learn in this text. It may help if you remember that the quotient rule resembles the Product Rule. ⎛ f (x) ⎞′ f ′(x) ⎜ g(x) ⎟ ≠ g′(x) ⎝ ⎠

Example

30

Solution

Also note that like in the Product Rule, the derivative of a quotient is not equal to the quotient of derivatives.

Find the derivative of the function f (x) = Using the Quotient Rule: f ′(x) =

(3x+ 1)′ (2 x − 1) − (3 x+1) (2 x − 1)′ (2 x − 1)2

f ′(x) =

3 ⋅ (2 x − 1) − (3 x+1) ⋅ 2 6 x − 3 − 6 x − 2 = (2 x − 1)2 (2 x − 1)2

f ′(x) = −

Example

31

Solution

304

3x+ 1 . 2x − 1

5 . ( 2 x − 1)2

2 Differentiate the rational function f (x) = x + x − 21 . x −1

According to the Quotient Rule, f ′(x) =

(2 x+ 1) ⋅ (x − 1) − (x2 + x − 21) ⋅ 1 (x − 1)2

f ′(x) =

2 x2 − x − 1 − x2 − x+ 21 x2 − 2 x+ 20 = (x − 1)2 (x − 1)2

f ′(x) =

x2 − 2 x+ 20 . x2 − 2 x+ 1 Algebra 10

32

Example

Solution

Differentiate the function f (x) =

2 x2 + 3x+ 1 . 2x

Before trying to use the Quotient Rule let us simplify the formula of the function: f (x) =

2 x2 + 3x+ 1 2 x2 3x 1 3 1 = + + = x+ + x −1. 2x 2x 2x 2x 2 2

In this example, finding the derivative will be easier and quicker without using the Quotient Rule. f ′(x) = 1+ 0 −

1 −2 1 x2 − 1 x =1− 2 = 2 x x2

Note We do not need to use the Quotient Rule every time we differentiate a quotient. Sometimes performing division gives us an expression which is easier to differentiate than the quotient. Let us verify the Quotient Rule.

Proof

(Quotient Rule) Let Q(x) =

f (x) and Q(x) be differentiable. g(x)

We can write f(x) = Q(x)g(x). If we apply the Product Rule: f ′(x) = Q′(x)g(x) + Q(x)g′(x) Solving this equation for Q′(x), we get f ′(x) − Q(x)g ′(x) = Q′(x) = g(x) Q′(x) =

Example

33

Solution

f (x) ⋅ g ′(x) g(x) g(x)

f ′(x)g(x) − f (x)g ′(x) . ( g(x))2

f(x) = ñx ⋅ g(x), where g(4) = 2 and g′(4) = 3. Find f ′(4). f ′(x) = ( x ⋅ g( x)) ′ = ( x ) ′ ⋅ g( x)+ x ⋅ g ′( x) f ′(x) =

g(x) 2 x

So f ′(4) = Differentiation

f ′(x) −

+ x ⋅ g ′(x)

g(4) 2 4

+ 4 ⋅ g ′(4) =

2 13 +2 ⋅3 = . 2⋅2 2 305

Check Yourself 7 1. Find the derivative of each function using the Product or the Quotient Rule. a. f(x) = 2x(x2 + x + 1)

b. f(x) = (x3 – 1)(x2 – 2)

⎛ 1 ⎞⎛ 1 ⎞ c. f (x) = ⎜ 2 + x ⎟⎜ + 1⎟ ⎝x ⎠⎝ x ⎠

1 ⎞ ⎛ d. f (x) = ( x + 1) ⎜ x 2 + ⎟ x⎠ ⎝

e. f (x) =

2 x+ 4 3x − 1

f. f (x) =

g. f (x) =

x2 − x+ 10 x+ 1

h. f (x) =

x −1 x +1 x3 + 3x2 − 5x+ 6 2x

2. If f(x) is a differentiable function, find an expression for the derivative of each function. a. y = x2f(x)

b. y =

f (x) x2

2 c. y = x f (x)

d. y =

1+ xf (x) x

3. Suppose that f and g are two functions such that f(5) = 1, f ′(5) = 6, g(5) = –3 and g′(5) = 2. Find each value. ⎛ f ⎞′ b. ⎜ ⎟ (5) ⎝ g⎠

a. (fg)′(5)

⎛ g ⎞′ c. ⎜ ⎟ (5) ⎝f⎠

Answers 1. a. 6x2 + 4x + 2 e. –

14 (3x – 1)2

b. 5x4 – 6x2 – 2x f.

1 2

x( x +1)

c. 1 –

2 3 – 4 3 x x

2 g. x + 2 x −2 11 ( x +1)

d.

5 1 x x + 2x − 2 2x x

h. x −

2 2 ′ ′ 2. a. 2xf(x) + x2f ′(x) b. f ( x) ⋅ x 4– 2 xf ( x) c. 2 xf ( x) – f (2x) ⋅ x x ( f ( x))

3. a. –16 b. –

306

20 9

3 3 + x2 2 2 ′ d. x f ( x2 ) – 1 x

c. 20

Algebra 10

C. THE CHAIN RULE We have learned how to find the derivatives of expressions that involve the sum, difference, product or quotient of different powers of x. Now consider the function given below. h(x) = (x2 + x – 1)50 In order to differentiate h(x) using the rules we know, we need to expand h(x), then find the derivative of each separate term. This method is, however, tedious! Consider also the function m( x) = x2 + x − 1. This function is also difficult to differentiate using the rules we have learned. For each of the two functions h(x) and m(x), the differentiation formula we learned in the previous sections cannot be applied easily to calculate the derivatives h′(x) and m′(x). We know that both h and m are composite functions because both are built up from simpler functions. For example, the function h(x) = (x2 + x – 1)50 is built up from the two simpler functions f(x) = x50 and g(x) = x2 + x – 1 like this: h(x) = f(g(x)) = [g(x)]50 = (x2 + x – 1)50 Here we know how to differentiate both f and g, so it would be useful to have a rule that tells us how to find the derivative of h = f(g(x)) in terms of the derivatives of f and g. THE CHAIN RULE [ f ( g( x))]′ = f ′( g( x)) ⋅ g′( x ) outer function

argument of the outer function

derivative of the outer function

derivative of the argument of the outer function

For example, if h(x) = f(g(x)) = (x2 + x – 1)50, then h′(x) = f ′(g(x)) ⋅ g′(x) = 50(x2 + x – 1)49 ⋅ (2x + 1).

1. [(x2 + x – 1)50]′ ≠ 50(x2 + x – 1)49 2. [(x2 + x – 1)50]′ ≠ 50(2x + 1)49 Differentiation

307

Example

34

Solution Example

35

Solution

Differentiate the function f(x) = (3x2 + 5x)2005. 2 2004 ⋅ (3 x2 +5 x) ′= 2005(3 x2 +5 x) 2004 ⋅(6 x+5). By the Chain Rule, f ′( x) = 2005(3 x +5 x)

Suppose m(x) = f(g(x)) and g(1) = 5, g ′(1) = 2, f(5) = 3 and f ′(5) = 4 are given. Find m ′(1). By the Chain Rule, m′(x) = f′(g(x)) ⋅ g ′(x). So m ′(1) = f ′(g(1)) ⋅ g′(1) = f ′(5) ⋅ 2 = 4 ⋅ 2 = 8.

Note The Chain Rule can be generalized for the composition of more than two functions as follows: [ f1( f2( f3(...fn(x)...)))]′ = [ f1′( f2( f3(...fn(x)...)))] ⋅ ( f2′( f3(...fn(x)...))) ⋅ ( f3′(...fn(x)...)) ⋅ ... ⋅ fn′(x) Using the Chain Rule we can generalize the Power Rule as follows: GENERAL POWER RULE [(f(x))n]′ = n(f(x))n – 1 ⋅ f ′(x) By using this rule we can more easily differentiate the functions that can be written as the power of any other functions. Example

36

Solution

Differentiate the function m( x) = x2 + x − 1. 1

We can rewrite the function as m(x) = (x2 + x – 1)2 and apply the General Power Rule: 1 – 1 m′( x) = ( x2 + x − 1) 2 ⋅ ( x2 + x − 1) ′ 2 1 – 1 m′( x) = ( x2 + x − 1) 2 ⋅ (2 x +1) 2

m′( x) =

Example

37

Solution

2 x +1 2 x2 + x − 1

Differentiate the function f ( x) =

1 . x + 3x 2

f ′( x) =[( x2 + 3x)−1]′ = −1( x2 + 3 x) −2 ⋅( x2 + 3 x) ′ f ′( x) = −1( x2 + 3 x) −2 ⋅ (2 x + 3) f ′( x) = −

308

2x + 3 ( x2 + 3x)2 Algebra 10

Example

38

Solution

–

4 – 1 f ′( x) = − (2 x3 + x2 − 15) 3 ⋅(2 x3 + x2 −15) ′ 3

f ′( x) = − f ′( x) = −

Example

39

Solution

1

Differentiate the function f(x) = (2x3 + x2 – 15) 3 .

1 1 ⋅ ⋅ (6 x2 + 2 x) 3 3 3 (2 x + x2 − 15)4 6 x2 + 2 x 3 3 (2 x3 + x2 − 15)4 −

2

−3 Differentiate the function f ( x) = (( x +1) 3 +5 x) .

−

2

−

2

f ′( x) = −3(( x +1) 3 +5 x) −4 ⋅ (( x+1) 3 +5 x) ′ 2 5 − − 2 f ′( x) = −3(( x +1) 3 +5 x) −4 ⋅ ( − ( x+1) 3 ⋅ ( x+1) ′+5) 3 2 5 − − 2 f ′( x) = −3(( x +1) 3 +5 x) −4 ⋅ (5 − ( x+1) 3 ) 3

Example

40

Solution

Differentiate the function f ( x) = (2 x − 3)5 ⋅ x2 − 2 x. The function is the product of two expressions, so we can use the Product Rule: f ′( x) = ((2 x – 3) 5 )′ ⋅ x2 − 2 x +(2 x − 3) 5 ⋅ ( x2 − 2 x) ′ f ′( x) = 5 ⋅ (2 x − 3)4 ⋅ 2 ⋅ x2 − 2 x +(2 x − 3) 5 ⋅ f ′( x) =10(2 x − 3) 4 x2 − 2 x +

Differentiation

1 – 1 ⋅( x2 − 2 x) 2 ⋅(2 x – 2) 2

(2 x − 3)5 ⋅ (2 x − 2) 2 x2 − 2 x

309

Example

41

Solution

7

⎛ 2t +1 ⎞ Differentiate the function g(t ) = ⎜ ⎟ . ⎝ t−3 ⎠ 6 ⎛ 2t +1 ⎞ ⎛ 2t +1 ⎞′ g′(t ) = 7 ⎜ ⎟ ⋅⎜ ⎟ ⎝ t−3 ⎠ ⎝ t −3 ⎠

(by the Power Rule)

6

⎛ 2t +1 ⎞ 2 ⋅ ( t − 3) − 1 ⋅ (2 t +1) g′(t ) = 7 ⎜ ⎟ ⋅ (t − 3)2 ⎝ t−3 ⎠

(by the Quotient Rule)

6

⎛ 2t +1 ⎞ 2t − 6 − 2t − 1 g′(t ) = 7 ⎜ ⎟ ⋅ (t − 3)2 ⎝ t−3 ⎠ g′(t ) =

(simplify)

−49(2t +1)6 . (t − 3)8

Notation

dy . dx If y = f(g(x)) such that y = f(u) and u = g(x), then we can denote the derivative of f(g(x)) dy dy du = ⋅ . by y′ = f ′(g(x)) ⋅ g′(x) or y′ = f ′(u) ⋅ u′(x) or dx du dx

Remember that if y = f(x), then we can denote its derivative by y′ or

The notation

Example

42

Solution

dy dy du is called Leibniz notation for the Chain Rule. = ⋅ dx du dx

Given that y = u2 – 1 and u = 3x2 + 1, find

dy by using the Chain Rule. dx

By the Chain Rule, dy dy du = ⋅ dx du dx dy d d = ( u 2 – 1) ⋅ (3 x+1) (find the derivative of the first function with respect to u dx du dx and the second function with respect to x) dy = (2 u – 1) ⋅ 3 dx dy = (2 ⋅ (3 x+1) – 1) ⋅ 3 dx dy =18 x + 3. dx

310

Algebra 10

Check Yourself 8 1. Find the derivative of f(x) = (2x + 1)3. 2. Differentiate y = (x3 – 1)100. 3. Find f ′(x) given f ( x) =

1 3

2

x + x +1

4. Find the derivative of g( x) = 4 5. y =

.

x3 − 1 . x3 +1

1 dy and u = 3x – 1 are given. Find . u dx

Answers 1. 6(2x + 1)2

4.

Differentiation

1 x3 +1 43 6 x2 ( ) 4 x3 – 1 ( x3 +1)2

2. 300x2(x3 – 1)99

5. –

3.

2 x +1 3 ( x2 + x +1)4 3

3 (3x – 1)2

311

D. HIGHER ORDER DERIVATIVES If f is a differentiable function, then its derivative f ′ is also a function, so f ′ may have a derivative of its own, denoted by (f ′)′ = f ′′. This new function f ′′ is called the second derivative of f because it is the derivative of the derivative of f. Look at three different ways of writing the second derivative of a function: d2 y f ′′( x) = y′′ = 2 dx Example

43

Solution

Find the second derivative of the function f ( x) =

2x . x –1

By the Quotient Rule, f ′( x) =

2 ⋅ ( x – 1) – 2 x ⋅ 1 2 x – 2 – 2 x 2 = =– . 2 2 ( x – 1) ( x – 1) ( x – 1) 2

Now differentiate f ′(x) to get f ′′(x): 2 ⎛ f ′′( x) = ⎜ – 2 ⎝ ( x – 1)

4 ⎞′ –2 –3 ⎟ = (–2( x – 1) ) ′ = 4( x – 1) ⋅1= ( x – 1)3 ⎠

Note Before computing the second derivative of a function, always try to simplify the first derivative as much as possible. Otherwise the computation of the second derivative will be more tedious. Notation

If we differentiate the second derivative f ′′(x) of a function f(x) one more time, we get the third derivative f ′′′(x). Differentiate again and we get the fourth derivative, which we write as f (4)(x) since the prime notation f ′′′′(x) begins to get difficult to read. In general, the derivative obtained from f(x) after n successive differentiations is called the nth derivative or dn y the derivative of order n and written by f (n)(x) or . dxn Example

44

Solution

Find the derivatives of all orders of the polynomial function f(x) = x5 + 4x4 + 2x3 – 5x2 – 6x + 7. f ′(x) = 5x4 + 16x3 + 6x2 – 10x – 6 f ′′(x) = 20x3 + 48x2 + 12x – 10 f ′′′(x) = 60x2 + 96x + 12 f (4)(x) = 120x + 96 f (5)(x) = 120 f (n)(x) = 0

312

(for n > 5) Algebra 10

Example

45

Solution

1 Find a general expression for the nth derivative of the function f ( x) = . x dy 1 = ( x–1 )′ = – x–2 = – 2 dx x d2 y 2 = (– x–2 )′ = 2 x–3 = 3 2 dx x 6 d3 y = (2 x–3 )′ = –6 x–4 = – 4 3 dx x 24 d4 y = (–6 x–4 )′ = 24 x–5 = 5 dx4 x 120 d5 y = (24 x–5 )′ = –120 x–6 = – 6 dx5 x

n!=n⋅(n–1)⋅(n–2)⋅...⋅3⋅2⋅1 for any natural number n

(–1)n n! dn y n –( n+1) ′ ⋅ = (...) = (–1) ! = n x dxn xn+1

Check Yourself 9 1. Find the second derivative of each function. a. f(x) = x3 – 3x2 + 4x + 5

b.

f ( x) =

x –1 x+ 2

2. Find the third derivative of each function. a. f(x) = x

2/3

b.

1 f (t ) = ( t 2 – 1)5 2

Answers 1. a. 6x – 6

Differentiation

b. –

6 ( x + 2)3

7 − 2. a. 8 x 3 27

2 2 2 b. 15t( t − 2) (3t − 2) 2

313

Motion is one of the key subjects in physics. We define many concepts and quantities to explain the motion in one dimension. We use some formulas to state the relations between the quantities. Derivative plays an important role in defining the quantities and producing the formulas from other derivatives. Here we will give the definition of important concepts and formulas that includes the uses of derivative. DISPLACEMENT The displacement is the change in the position of an object. If we denote the position at time t1 by x1, and the position at time t2 by x2, then the displacement is the difference between these two points; this is defined by Δx = x2 – x1. The time interval is, similarly, Δt = t2 – t1. We use the capital Greek letter Δ(delta) to show a change in a variable from one value to another. VELOCITY The velocity describes how fast the position of an object changes. It is measured over a certain time interval. If a car has a displacement Δx in a particular time interval Δt, then the car's average velocity, Vav, over that time interval is defined by Vav =

displacement x2 − x1 Δx = . = time interval t2 − t1 Δt

The definition of the average velocity includes a time interval. We learn more about the motion when smaller time intervals are used. Because of this, we define the instantaneous velocity as follows. The instantaneous velocity at a time t is the velocity of an object at that given instant of time. In other words, it is the limit of the average velocity as Δt approaches zero:

x(t + Δt ) − x(t ) Δt From the definition given above, we may conclude that the instantaneous velocity is the derivative of the displacement with respect to time t. V(t )= lim

Δt → 0

V (t )= lim

Δt → 0

Δx dx = Δt dt

ACCELERATION The term acceleration refers to the rate of change in velocity of an object with respect to time. We define the average acceleration, aav, in terms of velocity v1 at time t1 and v2 at time t2: v2 − v1 Δv = t2 − t1 Δt Now we will define the instantaneous acceleration as follows. aav =

Δv The instantaneous acceleration is the limit of the expression as the time interval goes to Δt zero. Δv dv = Δt dt This means that the instantaneous acceleration is the derivative of velocity with respect to time. Also it is the second derivative of the displacement. a(t ) lim = Δt → 0

For example, the position function x(t) of a car moving along a straight line is given as x(t) = 4t2 + 6t – 20 m where t is in seconds. The derivative of the position function gives the velocity function. V (t )=

The acceleration is

dx =(4t 2 +6t − 20)′ = 8t +6 m/s dt

dv =(8t +6)′ = 8 m/s 2. dt So, the car moves with constant acceleration. a(t )=

EXERCISES

4 .2

A. Basic Differentiation Rules 1. Find the derivative of each function by using the rules of differentiation.

c. f(x) = 10(3x + 1)(1 – 5x) d. f(x) = (x3 – 1)(x + 1)

c. f(x) = e f ( x) =

Product or the Quotient Rule.

b. f(x) = (2x + 3)(3x – 4)

1 f (x ) = − 151 π

d.

2. Find the derivative of each function by using the a. f(x) = 5x(x2 – 1)

a. f(x) = ñ2 b.

B. The Product and The Quotient Rules

e. f(x) = (x3 – x2 + x – 1)(x2 + 2) f. f(x) = (1 + ñt )(2t2 – 3)

1 8 x 12

f ( x) =

3 2x + 4

h. f ( x) =

x −1 2 x +1

i.

f ( x) =

1 − 2x 1+ 3x

j.

f ( x) =

x x2 +1

k.

f ( x) =

x2 + 2 x2 + x +1

l.

f ( x) =

x + 3x 3x − 1

g.

e. f(x) = 2x0.8 f. g.

f ( x) = f ( x) =

5 4/5 x 4 2 6

411

h. f(x) = 0.3x0.7 i. f(x) = 7x–12 j. f(x) = 5x2 – 3x + 7 x 3 + 2 x2 + x − 1 x

k.

f ( x) =

l.

4 3 2 f ( x) = 4 − 3 + t t t

o. 316

f ( x) =1 −

2 x

+

1 3 + x x

g′(1) = 3, find the value of h′(1). a. h(x) = f(x) ⋅ g(x)

m. f ( x) = x + 3 x + 5 x n. f ( x) = x +

3. Given that f(1) = 2, f ′(1) = –1, g(1) = –2 and

1 x

b. h(x) = (x2 + 1) ⋅ g(x) c. h( x) =

xf ( x) x + g( x)

d. h( x) =

f ( x) ⋅ g( x) f ( x) − g( x ) Algebra 10

4. Differentiate the function f ( x) =

x − 3x x

by

x simplifying and by the Quotient Rule. Show that

7. h(x) = g(f(x)) and f(2) = 3, f ′(2) = –3, g(3) = 5 and g′(3) = 4 are given. Find h′(2).

both of your answers are equivalent. Which method do you prefer? Why?

5. f(3) = 4, g(3) = 2, f ′(3) = –6 and g′(3) = 5 are given. Find the value of the following expressions. a. (f + g)′(3)

8. By using the Chain Rule, find

dy for each function. dx

a. y = u2 – 1, u = 2x + 1 b. y = u2 + 2u + 2, u = x – 1

b. (fg)′(3)

⎛ f ⎞′ c. ⎜ ⎟ (3) ⎝ g⎠

⎛ f ⎞ d. ⎜ ⎟ (3) ⎝ f − g⎠

c. y =

1 , u = x3 u –1

d. y = u +

1 u

, u = x2 – x

C. The Chain Rule 6. Find the derivative of each function. a. f(x) = (3x – 1)2 b. f(x) = (x2 + 2)5

D. Higher Order Derivatives 9. Find the second derivative of each function.

c. f(x) = (x5 – 3x2 + 6)7

a. f(x) = 3x2 – 7x + 2

d. f(x) = (x – 2)–3 2 e. f ( x) = (5x2 + 3x – 1)2

b. f(x) = (x2 + 1)7

f. f ( x) =

c. f ( x) =

1 4x2 +1

x2 x –1

d. f ( x) = 2 x – 1

g. f ( x) = ( x +1+ x ) 3 › h. f(x) = (x – 1)5 ⋅ (3x + 1)1/3 › i.

(1 – 3 x)7 f ( x) = (2 x +1)4

› j.

f ( x) = (

3x – 9 3 ) 2x + 4

2x – 1 3x +1 › l. f(x) = 3x + [2x2 + (x3 + 1)2]3/4 › k. f ( x) =

Differentiation

10. Find the third derivative of each function. a. f(x) = 5x4 – 3x3 b. f(x) =

2 x

c. f ( x) = 3x – 2 d. f(x) = (2x – 3)4 317

Mixed Problems

16. The concentration of a certain drug in a patient’s bloodstream t hours after injection is given by

11. Find the equation of the tangent line to the graph

0.2t . t2 +1 a. Find the rate at which the concentration of C(t ) =

of the function f(x) = (x3 + 1)(3x2 – 4x + 2) at the point (1, 2).

the drug is changing with respect to time.

x 12. The curve y = 2 is called a serpentine x +1 curve. Find the equation of the tangent line to the

b. How fast is the concentration changing 1 hour after the injection? What about after 2 hours?

curve at the point x = 3.

13. f is a differentiable function. Find an expression for the derivative of each of the following functions. a. y = x2ñxf(x) b. y = x3(f(x))2 c. y = d. y =

x3 f ( x) x + xf ( x ) x

14. Prove that (fgh)′ = f ′gh + fg′h + fgh′ if f, g and h ›

are differentiable functions.

17. g(x) = f(x2 + 1) is given. Find g ′(1) if f ′(2) = 3. ›

18. Find an expression for the derivative of ›

15. A scientist adds a toxin to a colony of bacteria. He

⎛ g( x)h( x) ⎞ f⎜ ⎟ if f, g, h, m and n are differentiable ⎝ m( n( x)) ⎠ functions.

estimates that the population of the colony after t 24t +10 hours will be P(t ) = 2 thousand bacteria. t +1 Find the estimated rate of change of the

19. If the tangent to the graph of f at point (2, 3) has an

population after three hours.

20. Given that f ( x) = x x x and f ′′( a) = − 7 ,

›

›

angle of 60° with x-axis, find the slope of tangent to the graph of g(x) = f 2(x) – x ⋅ f(x) at x = 2.

64

find a.

21. Given that f(4 ⋅ g(x) + 7) = x3 – 2x2 + 3 and ›

g(x) = 1 – x, find f ′(–1).

22. Find an expression for the nth derivative of the ›

318

function f ( x) =

1 . 2x Algebra 10

A. DERIVATIVES OF TRIGONOMETRIC FUNCTIONS Let us begin by looking at the derivatives of the sine and cosine functions. DERIVATIVES OF SINE AND COSINE (sin x)′ = cos x (cos x)′ = –sin x

46

Example

Solution

Find the derivative of the function f(x) = (sin x + cos x)2. f ′( x) = 2(sin x + cos x)(sin x+ cos x) ′

(by the General Power Rule)

f ′( x) = 2(sin x + cos x)(cos x – sin x)

(by the sum, derivative of the sine and cosine)

f ′( x) = 2(cos x – sin x)

(simplify)

f ′( x) = 2 cos 2 x

(by the trigonometric identity)

2

cos 2x = cos2 x – sin2 x

2

Now let us derive the formula for the derivative of the function f(x) = sin x. Proof

(Derivative of Sine Function) By the definition of the derivative, we have f ′( x) = lim

f ( x + h) − f ( x ) sin( x + h) − sin x = lim → h 0 h h

f ′( x) = lim

sin x cos h + cos x sin h − sin x sin x cos h − sin x cos x sin h ⎞ = lim ⎛⎜ + ⎟ h→ 0 ⎝ h ⎠ h h

h→ 0

h→ 0

lim h→0

sin h =1 h

cos h − 1 lim =0 h→0 h

Example

47

Solution

⎛ cos h − 1 sin h ⎛ cos h − 1 ⎞ ⎛ sin h ⎞ ⎞ sin x ⋅ lim + l im cos x ⋅ lim f ′( x) = lim ⎜ sin x ⎜ ⎟ + cos x ⎜ ⎟ ⎟ = lim → → → h→ 0 h 0 h 0 h 0 h → 0 h h h ⎝ ⎠ ⎝ h ⎠⎠ ⎝ f ′( x) = sin x ⋅ 0 + cos x ⋅ 1 = c os x.

Find the derivative of the function f(x) = x ⋅ sin x. By the Product Rule f ′( x) = ( x ⋅ sin x) ′ = ( x) ′ ⋅ sin x+ x ⋅(sin x) ′= sin x+ xcos x.

Differentiation

319

Example

48

Solution

Example

49

Solution

Find the derivative of the function f(x) = cos(x3 – x). f ′( x) = (cos( x3 − x)) ′ = − sin( x3 − x) ⋅ ( x3 − x) ′= −sin( x3 − x) ⋅(3 x2 −1)

Find the derivative of the function f(x) = sin3x2. In this example we have the composition of three functions. f ( x) = sin 5 x2 = (sin( x2 ))5 We apply the Chain Rule beginning from the outermost function: f ′( x) = ( (sin( x2 ))5 ) ′ = 5(sin( x2 ))4 ⋅ ( sin( x2 )) ′

sin 2x = 2sin x cos x

f ′( x) = 5(sin( x2 ) )4 ⋅ cos( x2 ) ⋅ ( x2 ) ′ f ′( x) = 5(sin( x2 ) )4 ⋅ cos( x2 ) ⋅ 2 x f ′( x) =10 x si n( x2 )cos( x2 )

Check Yourself 10 Find the derivative of each function. 1. f(x) = x – 3 sin x

2.

f(x) = xcos x

3. f ( x) =

sin x 1+ cos x

4. f(x) = cos2(x2 + x – 1) Answers 1. 1 – 3cos x

2. cos x – xsin x

4. –sin(2x2 + 2x – 2) ⋅ (2x + 1)

3.

1 1+ cos x

DERIVATIVES OF OTHER TRIGONOMETRIC FUNCTIONS (tan x)′ = sec2 x = 1 + tan2 x

(tan f(x))′ = sec2 f(x) ⋅ f ′(x)

(cot x)′ = –csc2 x = –(1 + cot2 x) sin x (sec x)′ = sec x ⋅ tan x = cos2 x

(cot f(x))′ = –csc2 f(x) ⋅ f ′(x)

(csc x)′ = –csc x ⋅ cot x = − 320

cos x sin2 x

(sec f(x))′ = sec f(x) ⋅ tan f (x) ⋅ f ′(x) (csc f(x))′ = –csc f(x) ⋅ cot f(x) ⋅ f ′(x) Algebra 10

Example

50

Solution

Find the derivative of the function f(x) = tan(x2 – 3x + 1). f ′(x) = sec2(x2 – 3x + 1) ⋅ (x2 – 3x + 1)′ = sec2(x2 – 3x + 1) ⋅ (2x – 3) or

Example

51

Solution

= (1 + tan2(x2 – 3x + 1))(2x – 3).

Find the derivative of the function f ( x) =

sec x . 1+ tan x

By the Quotient Rule, f ′( x) =

(sec x)′ ⋅ (1+ tan x) − sec x ⋅(1+ tan x) ′ (by the Quotient Rule) (1+ tan x)2

f ′( x) =

sec x tan x ⋅ (1+ tan x) − sec x ⋅ sec 2 x (1+ tan x)2

(differentiate)

f ′( x) =

sec x(tan x + tan 2 x − sec 2 x) (1+ tan x)2

(factorize)

f ′( x) =

sec x(tan x − 1) (1+ tan x)2

(simplify using tan2 x + 1 = sec2 x)

Check Yourself 11 Find the derivative of each function. tan x 1. f ( x) = 2. f(x) = 4 sec x – cot x x

3. f(x) = cot(x2 – x + 1)

Answers 1.

Differentiation

x sec 2 x – tan x x2

2. 4sec xtan x + csc2 x

3. csc2(x2 – x + 1) ⋅ (1 – 2x)

321

B. DERIVATIVES OF SPECIAL FUNCTIONS 1. Absolute Value Functions

y

⎧⎪ g′( x), g( x) > 0 If f(x) = |g(x)|, then f ′( x) = ⎨ . ⎪⎩ − g′( x), g( x) < 0 In general, a derivative does not exist when the function has 0 as

y = |x|

value because of a ‘corner’ as demonstrated in the graph on the right.

corner

x

When g(x) = 0, the derivative exists only when the right-hand side and the left-hand side derivatives are equal to each other. We can also formulate the derivative expression as follows: f ′( x) =

Example

52

Solution

| g( x)| ⋅ g ′( x) = sgn[ g( x)] ⋅ g ′( x), g( x) ≠ 0. g( x)

Find the derivative of the function f(x) = |1 – x2| at the points x = 2, x = 1 and x = 0. Let us begin by trying to rewrite the function as a piecewise function. The roots of the function are x = –1 and x = 2.

A piecewise function is a function that is defined by different formulae in different parts of its domain.

2 ⎪⎧ x − 1, x < −1 and x ≥ 1 Then the function f will be f ( x) = ⎨ . 2 ⎪⎩1 − x , –1 ≤ x < 1 Let us find f ′(2):

y

y = f(x)

Note that f(2) ≠ 0. 1

For x = 2, f ′(x) = (x2 – 1)′ = 2x. So, f ′(2) = 2 ⋅ 2 = 4. Let us find f ′(1):

–1

1

x

Since f(1) = 0, we will check the left-hand and the right-hand derivatives. For x < 1, f ′(x) = (1 – x2)′ = –2x. So, f ′(1–) = –2 ⋅ (1) = –2. For x > 1, f ′(x) = (1 – x2)′ = 2x. So, f ′(1+) = 2 ⋅ 1 = 2. Since f ′(1–) ≠ f ′(1+), f ′(1) does not exist. Let us find f ′(0): Note that f(0) ≠ 0. For x = 0, f ′(x) = (1 – x2)′ = –2x. So, f ′(0) = –2 ⋅ 0 = 0. 322

Algebra 10

Example

53

Solution

Given that f(x) = |x3 – 4x2 + 4x|, find the derivative of f(x) at the point x = 2. 3 2 ⎧− ⎪ x + 4x − 4x,x < 0 The piecewise form of the function is f ( x) = ⎨ . 3 2 ⎪⎩ x − 4x + 4x, x ≥ 0 Since f(2) = 0, we will check the left-hand and the right-hand derivatives.

For 0 ≤ x < 2,

f ′(x) = (x3 – 4x2 + 4x)′ = 3x2 – 8x + 4 f ′(2–) = 3 ⋅ (2)2 – 8 ⋅ 2 + 4 = 0.

For x > 2,

f ′(x) = (x3 – 4x2 + 4x)′ = 3x2 – 8x + 4 f ′(2+) = 3 ⋅ (2)2 – 8 ⋅ 2 + 4 = 0.

Since the left-hand and the right-hand derivatives are equal to each other, the derivative of the function exists at the point x = 2 and f ′(2) = 0. Example

54

Solution

Example

55

Solution

Given that f(x) = |x – x2|, find f ′(2) and f ′(3). Since f(2) ≠ 0 and f(3) ≠ 0, we can use the formula f ′( x) = f ′(2) =

| 2 − 22 | ⋅ (1 − 2 ⋅ 2) = −1 ⋅( −3) = 3 2 − 22

f ′(3) =

| 3 − 32 | ⋅ (1 − 2 ⋅ 3) = −1 ⋅( −5) = 5 3 − 32

| x − x2 | ⋅ (1 − 2 x). x − x2

Given that f(x) = |cos x|, find the derivative of f(x) at the points x =

For x =

π and x = π. 3

π , cos x > 0. So, f(x) = cos x and f ′(x) = – sin x. 3

π π 3 f ′( ) = − sin( ) = − 3 3 2

For x = π, cos x < 0. So, f(x) = –cos x and f ′(x) = sin x. f ′(π) = sin π = 0 Example

56

Solution

Given that f(x) = |x3 – 9| + x2, find f ′′(2). For x = 2, x3 – 9 < 0 and so f(x) = –x3 + 9 + x2. If we take the derivative twice, f ′(x) = –3x2 + 2x f ′′(x) = –6x + 2. Therefore, f ′′(2) = –6 ⋅ 2 + 2 = –10.

Differentiation

323

Check Yourself 12 1. Given that f(x) = |x2 – 3x|, find f ′(3) and f ′(5). 2. Given that f(x) = |x4 – 2x2 + 1|, find the derivative of f(x) at the point x = 1. π π 3. Given that f(x) = |sin x|, find f ′( ) and f ′( ) . 6 2 Answers 1. does not exist; 7 2. 0 3. 3 ; 0 2

2. Sign Functions Note that a sign function has the range {–1, 0, 1}. When it takes –1 and 1 as its value, the graph is a horizontal line. Since the slope of a horizontal line is 0, we have 0 as the derivative. When the function takes 0 as its value, the graph has a discontinuity. So, the derivative does not exist. Look at the graph on the right: In conclusion, if f(x) = sgn (g(x)), then ⎪⎧0, f ′( x) = ⎨ ⎪⎩does not exist,

Example

57

Solution

g( x) ≠ 0 g( x) = 0

y

slope = 0 1

y = sgn x

x –1

no slope

slope = 0

.

Given that f(x) = sgn(x2 – x), find the derivative of f(x) at the points x = –2 and x = 1. We begin by finding the value of f(–2) and f(1): f(–2) = sgn((–2)2 – (–2)) = sgn(6) = 1 f(1) = sgn(12 – 1) = sgn(0) = 0. Since f(–2) ≠ 0, f ′(–2) = 0. Since f(1) = 0, f ′(1) does not exist (f(x) is not continuous at x = 1).

324

Algebra 10

Example

58

Solution

Find the largest interval on which the function f(x) = sgn(x2 – x – 6) is differentiable. We can rewrite the function as a piecewise function: ⎧ 1, x < −2 and x > 3 ⎪ f ( x) = ⎨ −1, −2 < x < 3 . ⎪ ⎩ 0, x = −2 and x = 3

Since f(x) is not continuous at the points x = –2 and x = 3, it is not differentiable. It has a derivative at all other points, and this is equal to zero.

y 1

–2

y = f(x)

3 x

So, the largest interval on which f is differentiable is \ \ {–2, 3}. –1

3. Floor Functions ⎧⎪0, If f(x) = ag(x)b, then f ′( x) = ⎨ ⎪⎩ may not exist,

g( x) ∉ ] g( x) ∈ ]

.

When g(x) ∈ ], f(x) is certainly continuous and differentiable. However, when g(x) ∉ ], we cannot be certain. It may be differentiable or not. In order to determine whether a floor function is differentiable or not at a given value, we check the left-hand and the right-hand derivatives. Example

59

Solution

Given that f ( x) = a 2 x + 1b , find the derivative of f(x) at the points x = 2 and x = 4. 3 2 x +1 5 2x + 1 is not an integer for x = 2, f ′(2) = 0. = ∉ ]. Since the expression 3 3 3 2 x +1 = 3 ∈ ]. Here we have to find the left-hand and the right-hand derivatives, For x = 4, 3 because the expression 2 x +1 is an integer for x = 4. 3 f ( x) − f (4) 2 −3 −1 f ′(4− ) = lim− = – = − =+∞ x→ 4 x−4 4 −4 0

For x = 2,

f ′(4+ ) = lim+ x→ 4

f ( x) − f (4) 3 −3 0 = + = =0 x−4 4 − 4 0+

Since f ′(4–) ≠ f ′(4+), f ′(4) does not exist. Differentiation

325

Example

60

Solution

π Given that f(x) = asin xb, find f ′( ) and f ′(π). 6

π 1 π sin( ) = ∉ ]. So, f ′( ) = 0. 6 2 6 At x = π, f(x) is not continuous (see the graph). So, it is

y 1

not differentiable. Thus the derivative of f(x) does not exist at this point.

Example

61

Solution

p

–p/2 p/2 –1

2p x y = f(x)

Given that f(x) = ax2b, find f ′(0). For x = 0, f(x) = 0 ∈ ]. So, we have to find the left-hand and the right-hand derivatives of the function f(x) = ax2b at the point x = 0. f ′(0 − ) = lim−

f ( x) − f (0) a x2 b − 0 0 = lim− = − =0 x→0 x−0 x −0 0

f ′(0 + ) = lim+

f ( x) − f (0) ax2 b − 0 0 = lim+ = + =0 x→0 x−0 x −4 0

x→0

x→0

Notice that if –1 < x < 1, ax2b = 0. Since f ′(0–) = f ′(0+), the derivative of f(x) = ax2b exists at the point x = 0 and f ′(0) = 0.

Check Yourself 13 1. Given that f(x) = sgn (x2 + x), find f ′(–1) and f ′(2). π π 2. Given that f(x) = acos xb, find f ′( ) and f ′( ). 3 2 3. Given that f(x) = |x2 + 3x – 4| + ax2b + sgn (x2 – 1), find f ′(0). Answers 1. does not exist; 0 326

2. 0; does not exist

3. –3 Algebra 10

C. IMPLICIT DIFFERENTIATION Up to now we have worked with the functions expressed in the form y = f(x). In this form, the variable y is expressed easily in terms of the variable x. A function in this form is said to be in the explicit form. However, some functions cannot be written in explicit form. Consider the following equation: y5 + y + x = 0 If we are given a value of x, we can calculate y in this equation. However, we cannot write the equation in the form y = f(x). We say that x determines y implicitly, and that y is an implicit function of x. Look at the same more implicit functions: x5 + 2xy2 – 3y4 = 7 y – 2y2 = x x2 – y2 + 4y = 0 How can we differentiate an implicit function? Recall the Chain Rule for differentiation. In an implicit function, y is still a function of x, even if we cannot write this explicitly. So, we can use the Chain Rule to differentiate terms containing y as functions of x. For example, if we are differentiating in terms of x, dy (y4)′ = [(f(x))4]′ = 4( f(x))3f ′(x) = 4y3y′ or (y4)′ = y3 , dx dy (7y)′ = 7y′ or (7y)′ = 7 . dx The procedure we use for differentiating implicit functions is called implicit differentiation. Let us summarize the important steps involved in implicit differentiation. IMPLICIT DIFFERENTIATION 1. Differentiate both sides of the equation with respect to x. Remember that y is really a function of x and use the Chain Rule when differentiating terms containing y. dy 2. Solve the resulting equation for y′ or in terms of x and y. dx Example

62

Solution

Find y′ given the equation y5 + y + x = 0. ( y5 + y + x)′ = (0)′

(differentiate both sides)

( y5 )′ +( y)′+( x) ′ = 0

(by the Sum Rule)

5y4 y′ + y′ +1= 0

(by the Chain Rule)

y′(5y4 +1) = −1

(factorize)

y′ = − Differentiation

1 5y4 +1

(isolate y′ ) 327

Example

63

Solution

Find

( y3 − y2 x + x2 − 1)′ = (0)′

(differentiate both sides)

( y3 )′ − ( y2 x)′ + ( x2 )′ − (1)′ = 0

(by the Sum Rule)

3y2

Example

64

dy given the equation y3 – y2x + x2 – 1 = 0. dx

dy dy − (2 y x + y2 ) + 2 x − 0 = 0 dx dx

(by the Chain Rule and the Product Rule)

dy (3y2 − 2 yx) = y2 − 2 x dx

(factorize)

dy y2 – 2x = 2 dx 3y − 2 yx

(isolate

dy ) dx

The equation x2 + y2 = 4 is given. dy a. Find by implicit differentiation. dx b. Find the slope of the tangent line to the curve at the point (ñ3, 1). c. Find the equation of the tangent line at this point.

Solution

a. Differentiating both sides of the equation with respect to x, we obtain ( x2 + y2 )′ = (4) ′ ( x2 )′ +(y2 )′ = 0 2x + 2y

dy =0 dx

dy x =– dx y

( y ≠ 0).

b. The slope of the tangent line to the curve at the point (ñ3, 1) is given by dy dx ( a, b)

is used for slope of the curve at the point (a, b).

m=

dy dx (

=– 3, 1)

x y(

=– 3 , 1)

3 = – 3. 1

c. We can find the equation of the tangent line by using the point-slope form of the equation of a line. The slope is m = –ñ3 and the point is (ñ3, 1). Thus, y

y – y1 = m(x – x1) y – 1 = –ñ3(x – ñ3)

y + 3x – 4 = 0

ñ3x + y – 4 = 0. A sketch of this tangent line is given on the right. The line x + ñ3y – 4 = 0 is tangent to the graph of the equation x2 + y2 = 4 at the point (ñ3, 1). 328

2 –2

2 –2

4

x

x2 + y2 = 4

Algebra 10

Example

65

Solution

Find the derivative with respect to x of the implicit function

x2 + y2 + x2 = 2.

Differentiating both sides of the given equation with respect to x, we obtain d 2 d d ( x + y2 )1/ 2 + ( x2 ) = (2) dx dx dx 1 2 d 2 ( x + y2 )1/ 2 ( x + y2 ) + 2 x = 0 2 dx 1 2 dy ( x + y2 )–1/ 2 (2 x + 2 y )+ 2 x = 0 2 dx 2x+ 2y 2y

dy = –4 x( x2 + y2 )1/ 2 dx

dy = –4 x( x2 + y2 )1/ 2 – 2 x dx

2 2 dy –2 x x + y – x = . dx y

Check Yourself 14 dy by implicit differentiation. dx

1. Find

a. x3 + x2y + y2 = 5

b. x2y + xy2 = 3x

2. Find the equation of the tangent line to each curve at the given point. a. x2y3 – y2 + xy – 1 = 0; (1, 1) 2

2 2 b. x – y =1; (–5, 9 ) 16 9 4

2

c. x 3 + y 3 = 4; (1, 3 3) Answers –3x2 – 2 xy 1. a. x2 + 2 y 2. a. y = –

Differentiation

3 5 x+ 2 2

b.

3 – 2 xy – y2 x2 + 2 xy

b. y = –

5 x–4 4

c. y = –ñ3x + 4ñ3

329

D. DERIVATIVES OF PARAMETRIC FUNCTIONS Sometimes we express the variables x and y as functions of a third variable t by a pair of functions. x = f(t),

y = g(t)

Functions like these are called parametric functions, and the variable t is called the parameter. PARAMETRIC DIFFERENTIATION dy dy = dt dx dx dt

,

dx ≠0 dt

This enables us to find the derivative of a parametric function ( eliminate the parameter t. Example

66

Solution

Example

67

Solution

Example

dy ) without having to dx

Find the derivative with respect to x of the parametric curve x = t + 2 and y = 2t2 – 1. dy dy 4t = dt = = 4t dx dx 1 dt

If a is a positive constant and x = a cos t, y = a sin t, then find

dy . dx

dy dy a cos t = dt = = – cot t dx dx – a sin t dt

68

The parametric curve is given by the equations x = t +1 and y = t2 + 3t. Find the slope of its tangent at x = 2. dy Solution Let us begin by finding in terms of t. dx dy d 2 (t + 3t ) 2t + 3 dy = dt = dt = = 2( t +1)1/ 2 (2 t + 3) 1 dx d dx ( t +1) dt dt 2 t +1 For x = 2, x = t +1 = 2. So, t = 3. m=

330

dy dx

t=3

= 2 ⋅ ( t + 1)1/ 2 ⋅ (2t + 3) t=3 = 2 ⋅(3+1)1/ 2 ⋅(2 ⋅3+ 3) = 2 ⋅2 ⋅9 = 36. Algebra 10

PARAMETRIC DIFFERENTIATION OF SECOND ORDER dy′ d2 y dy′ = = dt dx dx2 dx dt

dy in terms of t. dx 2. Differentiate y′ with respect to t. dx . 3. Divide the result by dt

1. Express y′ =

Example

69

Solution

Find

d2 y , if x = 2t – t2 and y = 1 – t3. 2 dx

dy d (1 – t3 ) dy –3t 2 dy dt dt First, find in terms of t: = = = . d dx dx dx (2t – t 2 ) 2 – 2t dt dt

Then, differentiate

d dy d 3t 2 –6t ⋅ (2 – 2t) – (–3t 2 ) ⋅(–2) dy with respect to t: ( ) = (– )= . dt dx dt 2 – 2t dx (2 – 2t)2

Finally, divide the result by

6t 2 – 12t 3t dx dx d 2 y –12t +12t 2 – 6t 2 : : 2= = =– . 2 3 dt dt dx (2 – 2t) ⋅ (2 – 2 t) 8 ⋅(1 – t) 4( t – 1) 2

Check Yourself 15 dy for each parametric curve. dx b. x = 5cos t, y = 5sin t a. x = 2t + 3, y = t2 – 1

1. Find

d2 y for each parametric curve. dx2 dy a. x = 3t2 + 2, y = 2t2 – 1 b. = 4+ sin 2 t , x = cos 2 t dx Answers

c. x =

t –1 t +1 , y= t +1 t –1

2. Find

Differentiation

1. a. t

b. –cot t

2. a. 0

b. –

t +1 ⎞ c. − ⎛⎜ ⎟ ⎝ t −1⎠

2

1 4 4+ sin 2 t 331

E. DERIVATIVES OF INVERSE TRIGONOMETRIC FUNCTIONS In order to find the derivatives of the inverse trigonometric functions, we can use implicit differentiation. For example, what is the derivative of Arcsin x? π π Let y = Arcsinx, then sin y = x and − ≤ y ≤ . 2 2 Now, if we differentiate sin y = x implicitly with respect to x, we get cos y ⋅

dy dy 1 =1 or = . dx dx cos y

cos y ≥ 0, since −

1

π π ≤ y ≤ . So, cos y = 1 − sin 2 y = 1 − x2 . 2 2

x

y

d 1 1 Therefore, (Arcsin x) = = . cos y dx 1 − x2

1 – x2

We can use a similar method to find the derivative of the other inverse trigonometric functions.

DERIVATIVE OF INVERSE TRIGONOMETRIC FUNCTIONS 1

(Arcsin x)′ =

1− x

(Arccos x)′ = – (Arctan x)′ =

Example

70

Solution

332

1 1− x

1 − ( f ( x))2

(Arccos f( x)) ′= –

2

1 1+ x2

(Arccot x)′ = –

f ′( x)

(Arcsin f ( x)) ′=

2

(Arctan f ( x)) ′=

1 1+ x2

f ′( x) 1 − ( f ( x))2

f ′( x) ( f ( x))2 +1

(Arccot f( x)) ′= –

f ′( x) ( f ( x))2 +1

Find the derivative of the function f(x) = Arcsin (x2). f ′( x) = (Arcsin( x2 )) ′ =

( x2 )′ 2 2

1 − (x )

=

2x 1 − x4 Algebra 10

Example

71

Solution

Find the derivative of the function f(x) = x⋅ Arctan ñx. By the Product Rule, f ′( x) = ( x ⋅ Arctan x ) ′ =1 ⋅ Arctan x + x ⋅

Example

72

Solution

1

1 x . ⋅( x –1/ 2 ) = Arctan x + 2 2(1+ x) 1+ ( x ) 2

Find the equation of the tangent line to the curve f(x) = 2 Arccos

f ′( x) = –2 ⋅

⎛ x ⎞′ ⎜ ⎟ ⎝2⎠ x 1 – ⎛⎜ ⎞⎟ ⎝2⎠

2

= –2 ⋅

1 2 x 1 – ⎛⎜ ⎞⎟ ⎝2⎠

2

x at x = ñ3. 2

2

=–

4 – x2

The slope of the tangent line is f ′(ñ3 ) = –2. The curve passes through the point ( 3,

2π 2π ) since f ′(ñ3 ) = . 3 3

The equation of the tangent line is: y – 2π = –2( x – 3) or y = –2 x+ 2 3 + 2 π . 3 3

Check Yourself 16 1. Differentiate the functions. a. f(x) = (Arctan x)3 2. Find the equation of the tangent line to the curve y = Arccos 2x at the point where it crosses the y-axis. Answers 1. a.

Differentiation

3(Arctan x)2 1+ x2

2. y = –2 x +

π 2

333

EXERCISES

4 .3

A. Derivatives of Trigonometric Functions 1. Differentiate the functions. a. f(x) = sin (3x – 5)

B. Derivatives of Special Functions 4. Find the required values using the given data: ⎛ a. f ( x) =| 2 x − 3x2 |, f ′(0 − )+ f ′ ⎜ ⎛⎜ 2 ⎞⎟ ⎜⎝ 3 ⎠ ⎝

−

⎞ ⎟⎟+ f ′(1) = ? ⎠

c. f(x) = sin x – cos x

⎛ 1⎞ b. f ( x) = x2 + a xb + sgn( x – 2), f ′ ⎜ ⎟ = ? ⎝ 3⎠

d. f(x) = 2 tan x + sec x

c. f(x) = (2x + 3) ⋅ sgn(x2 + 1), f ′(ñ2) = ?

e. f(x) = sin x ⋅ tan x

d. f ( x) = sgn(4 x +5) ⋅a

b. f(x) = cos (x2 – 1)

f. f(x) = 2x tan x – x cos x

3x +1 b, f ′(2) = ? 2

g. f(x) = cos2(2x3 – 3x) h. f ( x) = ( i. f ( x) =

1 – cos x 10 ) 1+ cos x

4x , 2 x2 − 1 find the number of different x-values for which the

5. Given that f ( x) =| x2 − 4| − sgn( x3 + x)+

cot x 1+ sec x

j. f(x) = (1 + sec x) ⋅ (1 – cos x)

function is not differentiable.

k. f(x) = tan x2 – x – 1 l. f ( x) =

cot( x3 – 1) 1 – sec 2 ( x3 – 1)

m. f(x) = [x2sin(x – 1)]5 3 n. f ( x) = sec 2( 2x ) x –1

2. Find the equation of the tangent line to the curve at the given point. y = x cos x;

C. Implicit Differentiation dy for each equation below. dx a. 5x – 4y = 3 b. xy – y – 1 = 0 y c. x3 + x2 – xy = 1 d. – 3x2 = 5 x

6. Find

e. 2x2 + 3y2 = 12

f. x2 + 5xy + y3 = 11

g. x2y3 – xy = 8

h. òxy – 3x – y2 = 0

x=π

7. Find the equation of the tangent line to the given curve at the indicated point. 3. For what values of x does the graph of f(x) = x + 2 sin x have a horizontal tangent line? 334

a. 4x2 + 2y2 = 12; (1, –2) b. 2x2 + xy = 3y2; (–1, –1) Algebra 10

D. Derivatives of Parametric Functions dy 8 . Find for each pair of parametric equations. dx a. x = 3t – 1 and y = t2 – 2t 1 b. x = t(t + 1) and y = t – t c. x = t3 – t2 – 1 and y = t2 + 3t + 1

Mixed Problems 13. Find the given order derivative by finding the first few derivatives and observing the pattern that occurs. 27 a. d 27 (cos x) dx

b.

d99 (sin 2 x) dx99

35 c. d 35 ( x sin x) dx

d. x = t +1 and y = t2 + 3t e. x = 3 t and y = 4 – t 2 f. x = 4cos t and y = 5sin t

2 14. Find the second derivative d y2 of each implicit

9. Find the equation of the tangent line at the given point P. 1 a. x = + t 2 , y = t2 – t + 1, P(2, 1) t b. x = 3t2 + 2, y = 2t4 – 1, P(5, 1)

a. x = t2 – t y = t3 + 3t + 1

b. x = ñt + 1 1 y= +1 t

11 . Differentiate the functions. x a. f(x) = Arcsin 2 x2 – 1 b. f(x) = Arccos x 2 c. f(x) = Arctan (x + x – 1)

b. x3 + y4 = 20

c. y2 + xy = 8

d.

⎛π2 ⎞ f ( x) = sin x , f ′ ⎜ ⎟=? ⎝ 16 ⎠

17. Given that

π < x<π , 2

differentiate f ( x) =

normal line to y = Arcsin

›

Differentiation

x when x = 1. 2

|1 − tan x | ⋅ sgn(tan x) . cos x a b 2

18. Given that f(x) = x3|x2 – 2|, find f ′(–2) + f ′′(1).

19. Given that f ( x) =

12. Find the equation of the tangent line and the

x + 3 y =1

the curve y = x2 – 2|x – 1| at exactly two points.

d. f(x) = Arctan x – 1 – x2 e. f(x) = Arcsin (tan x)

3

16. Find the required values using the given data:

c. x = 2sin2 t y = 3cos2 t

E. Derivatives of Inverse Trigonometric Functions

a. x2y – 1 = 0

15. Write the equation of the line which is tangent to ›

d2 y 10. Find . dx2

dx

› function.

x2 +1 , solve f ′(x) > 0. x2 − 1

20. Given that f ( x) = (2 − x + 2 )2 , solve f ′(x) = 0. ›

335

CHAPTER REVIEW TEST

4A

1. f(x) = x3 – 2x2 + 1 is given. Find the value of the limit lim x→ 2

A) 2

f ( x) – f (2) . x–2

B) 3

C) 4

5. f(x) = sin x and g(x) = x2 + 3 are given. Find the derivative of f(g(x)).

D) 5x + 1

E) 3x2 – 4

A) 2x cos (x2 + 3)

B) 2x sin x

C) x + sin x

D) x sin (x2 + 3) E) –x cos (x2 + 3)

2. Find the equation of the tangent line to the π function f(x) = 3cos 2x – 1 at the point M( , − 1). 2

A) y = 1

B) y = 2

D) y = x – 2

C) y = –4 ⎧⎪ x3 , 6. f ( x) = ⎨ ⎪⎩3x,

E) y=–1

A) 0

3. The graph of the

B) 1

x ≤1 x >1

is given. Find f ′(1).

C) 2

D) 3

E) does not exist

y

parabola y = ax2 + bx + c is shown on the right such that t is its tangent at the point (1/2, –1). By using the information given in the figure, find the value of a + b + c. 1 A) –2 B) − C) 0 2

y = ax2 + bx + c t 1/2

45° x

–1

7. f(x) = 2x2 – 3x + 1 is given. Find f ′′(1). A) 8

D) 10

B) 6

C) 4

D) 3

E) 1

E) 15

8. f(x) = tan x – cot x is given. Find f ′(x). 4. f(x) = ax2 – 3x3 + 4 and f ′(2) = 8 are given. Find the value of a. A) 3 336

B) 4

C) 7

D) 11

E) 29

A)

4 sin2 2x

B)

D) tan2 x + cot2 x

3 sin 2x

C) 2tan2 x E) sin 2x Algebra 10

9. f ( x) = 2 x − 1 is given. Find f ′(5). A)

1 2

B)

1 3

C)

1 4

D)

x2 is tangent to the straight line a with the equation x – y = 1, then find the value

13. If the curve y = 1 5

E)

1 6

of a. A) 5

10. If the parametric function is given by the equations x = sin2 θ, y = sin2 θ, find A) 0

B) 1

C) –1

dy . dx

D) sin 2θ

E) –tan 2θ

B) 4

C) 3

dy . dt

B) 3t – 2t2x

A) 3 – 2t D) 3x + 2x2

C) 3x – 2x2t

E) 3t + 2x2t

expression f ′(1) + f ′(3). A) 0

B) 1

C) 2

the right belongs to the function f(x). f ( x) If g( x) = , then x find the slope of the

Chapter Review Test 4A

C) 15

y=

D) 18

y 1 1

2

x

4

y = f(x)

B) −

1 2

C) 2

D) 1

E) 0

16. Find the shortest distance between the curve

x −1

h(x) = x3 ⋅ f(x), find h′(1). B) 6

E) 4

graph of g(x) at the point x = 2.

12. Given that f(1) = 3, lim f ( x) − 3 = 6 and

A) 3

D) 3

tangent line to the

A) − 1 4

x →1

E) 1

14. If f(x) = |2 – x| + 2, then find the value of the

15. The graph given on

11. y = 3xt – x2t2 is given. Find

D) 2

E) 20

4 and the origin (0, 0). x

A) 8

B) 4

C) 2

D) 4ñ2

E) 2ñ2 337

CHAPTER REVIEW TEST

4B

1 2

1. f(x) = |x – 3| + sgn(x – 1) + a x + b is given.

5. Find lim x→0

Find the derivative of the function at the point A)

1 x= . 2

A) does not exist

B) 0

1 2

4

16+ x − 2 . x

B) 1 4

C) 1 8

D)

1 16

E)

1 32

C) 1

D) 2

E) 18

⎧ x2 + 2, x ≤ 1

6. f ( x) = ⎪⎨ 2. f ( x) = A) 9

3. Find

is given. ⎪⎩2 x +1, x > 1 Find the derivative of the function at x = 1.

3 x − is given. Find f ′(9). 3 x

B) 3

C) 1 3

D) 1 6

E) 1 9

d2 (sin x + cos x)2. dx2

A) 2(cos x – sin x) 2

C) sin x – cos x

B) 1

7. f ( x) = ( B) 2(sin x – cos x)

2

A) 0

D) 2 cos 2x

A) –300

C) 2

D) 3

E) does not exist

x+ 2 2 df (3) ) is given. Find . x−2 dx

B) –200

C) –150

D) –90

E) –40

E) –4 sin 2x

4. If the parametric curve is given by the equations x = t3 – 2t, y = t3 – 3t, find A) 20 338

B) 12

C) 6

d2 y dx2

. t =1

D) –20

E) –30

8. f(x) = |x2 + 3x – 4| + ax2b + sgn(x2 – 1) is given. At which one of the following points does the derivative of the function exist? A) 1

B) 0

C) –1

D) –2

E) –3 Algebra 10

9.

13. Which one of the following is correct for the

y M

2

tangent lines to the curve y =

l y = f(x)

–3

x

3

x = a and x = –a?

x3 at the points | x|

A) They are perpendicular to each other. B) They are parallel to each other. C) The angle between them is 30°. D) They are parallel to x-axis.

A is tangent to the curve y = f(x) at the point M(3, 2). If h( x) =

A)

2 9

E) They are parallel to y-axis.

f ( x) , find h′(3). x

B) −

5 9

C) −

1 9

D)

1 3

E) 4 3

1 df ( ) 4 . 14. f ( x) = (arcsin 2 x ) is given. Find dx 2

10. Find the slope of the normal line to the function f(x) = sin(cos 5x) at the point x =

π . 10

1 5

2 5

A) − 4 5

B) −

C)

1 5

D)

E) 4 5

A)

π2 4

B)

π2 2

C) π2

D) π

E) 2π

15. The tangent line to the curve y = x3 at the point A(2, 8) intersects the curve at another point B(x0, y0). Find x0.

11. f(x) = (x – 1)2 ⋅ (2x – t) and f ′′(0) = 0 are given.

A) −

3 2

B)

5 2

C) –3

D) –4

E) –5

Find the value of t. A) 4

B) 2

C) 0

D) –2

E) –4

16. The implicit function sin (xy) + cos (xy) = 0 is given. Find

12. f ( x) = 2+ x is given. Find f ′(4). A) 1

B) 4

Chapter Review Test 4B

1 C) 2

D) 1 4

dy . dx

A) 1 1 E) 16

B) − D) ycos (xy)

y x

C) 2xy E) xcos x – ycos y 339

APPLICATION SAYFASI EKLENECEK

340

Algebra 10

0 A. THE INDETERMINATE FORM – 0 Guillaume de L'Hospital (1661-1704)

French mathematician solved a difficult problem posed by Pascal at age 15. He published the first book ever on differential calculus “L'Analyse des Infiniment Petits pour l'Intelligence des Lignes Courbes” (1696). In this book, L'Hospital introduced L'Hospital's rule. Within the book, L'Hospital thanks Bernoulli for his help. An earlier letter by John Bernoulli gives both the rule and its proof, so it seems likely that it was Bernoulli who discovered the rule. L'Hospital’s name is spelled both “L'Hospital” and “L'Hôpital”, the two being equivalent in French spelling.

Let us consider the following limit where both f(x) and g(x) approach to zero as x → a: f ( x) lim x → a g( x) 0 If we substitute x = a in this limit, we obtain a fraction of the form , which is a 0 meaningless algebraic expression. This limit may or may not exist and is called an 0 indeterminate form . 0 From earlier studies you have learned to calculate such limits by using the limit theorems. In this section, we will discuss a very powerful method known as L’Hospital’s Rule. This rule 0 gives a connection between derivatives and limits of the indeterminate form . 0

L’HOSPITAL’S RULE Let the functions f and g be differentiable on an open interval that contains the point a. Suppose that lim f(x) = lim g(x) = 0 and lim x→a

x→ a

x→a

lim x→ a

EXAMPLE

1

Solution

Find lim x →0

f ′( x) exists. Then, g′( x)

f ( x) f ′( x) = lim = L. g( x) x→ a g′( x)

sin x . x

Since lim sin x = 0 and lim x = 0, we can apply L’Hospital’s Rule. x→0

x→0

sin x (sin x)′ cos x lim = lim = lim = cos 0 =1 x→0 x→0 x→0 x ( x)′ 1

Note Using L’Hospital’s Rule, differentiate both the numerator and the denominator seperately. Do not apply the Quotient Rule. 342

Algebra 10

EXAMPLE

2

Solution

Find lim x→ 4

x–4 . x2 – 4

If we attempted to use L’Hospital’s Rule, we would get x–4 ( x – 4)′ 1 1 lim 2 = lim 2 = lim = . x→ 4 x – 4 x → 4 ( x – 4)′ x→ 4 2 x 8 This is wrong! x–4 0 Since lim 2 does not give the indeterminate form , we cannot apply L’Hospital’s Rule x→ 4 x – 4 0 here. Let us find the correct answer: lim x→ 4

x–4 4–4 0 = 2 = =0 2 x – 4 4 – 4 12

Note Before applying L’Hospital’s Rule, verify that we have the indeterminate form

EXAMPLE

3

Solution

0 . 0

x3 + x + 2 . x → –1 x +1

Find lim

x3 + x + 2 (–1)3 +(–1)+ 2 0 = = x → –1 x +1 –1+1 0 lim

(

0 form; appl y the r ule) 0

x3 + x + 2 ( x3 + x + 2)′ 3 x2 +1 = lim = lim = 3 ⋅ (–1) 2 +1= 4 x → –1 x → –1 x → –1 x +1 ( x +1)′ 1 lim

EXAMPLE

4

Solution

Find lim x →1

lim x →1

ln x . x2 – 1

ln x 0 = x2 – 1 0

(

0 form; apply the rule ) 0

1 ln x (ln x)′ 1 lim 2 = lim 2 = l im x = x →1 x – 1 x →1 ( x – 1)′ x →1 2 x 2

Applications of the Derivative

343

EXAMPLE

5

Solution

Find lim

x →–1

2x + 3 – 1 x +5 – 2

.

0 . So, we can use L’Hospital’s Rule: 0 2 ′ 2x + 3 – 1 ( 2 x+ 3 – 1) 2 x+5 2 ⋅ 4 lim = lim = lim 2 2 x + 3 = lim = = 4. x → –1 x → –1 x →–1 1 x +5 – 2 ( x +5 – 2)′ x→–1 2 x+ 3 1 2 x +5

We have the indeterminate form

Note

f ′( x) 0 is still indeterminate form , we use L’Hospital’s Rule again. g′( x) 0 f ′( x) f ′′( x ) = lim . That gives lim x → a g′( x) x → a g ′′( x ) 0 In fact, whenever L’Hospital’s Rule gives , we can apply it again until we get a different result. 0

If lim x→ a

EXAMPLE

6

Solution

Find lim x→0

x – sin x . x2

lim

x – sin x 0 = x2 0

lim

x – sin x ( x – sin x)′ 1 – cos x 1 – cos 0 0 0 ( form; apply the rule agai n) = lim = lim = = 2 2 0 0 x → x → x ( x )′ 2x 2 ⋅0 0 0

lim

1 – cos x (1 – cos x) ′ sin x 0 = lim = li m = =0 x→0 x→0 2x (2 x)′ 2 2

x→0

x→0

x→0

(

0 form; apply the rule) 0

Check Yourself 1 Find the following limits: x3 – 8 x → 2 x2 – 4 Answers 1. 3 2. 6

1. lim

2. lim x→ 4

x2 – 8 x x–4

3. lim

x →π/2

sin x – 1 cos 2 x +1

3. –1/4

∞ B. THE INDETERMINATE FORM – ∞

∞ . It is expressed as follows: ∞ f ′( x) Suppose that lim f(x) = ±∞, lim g(x) = ±∞ and lim exists. x → a g′( x) x→a x→a

L’Hospital’s Rule is also valid for the indeterminate form

Then, lim x→ a

344

f ( x) f ′( x) = lim . x → a g( x) g′( x) Algebra 10

EXAMPLE

7

Solution

x2 – 3x +5 . x →∞ 7+ 2 x – 3 x2

Find lim

∞ form; apply the rule) ∞

lim

x2 – 3x +5 ∞ = x →∞ 7+ 2 x – 3 x2 –∞

(

x2 – 3x +5 ( x2 – 3 x+5) ′ 2x– 3 = lim = lim x →∞ 7+ 2 x – 3 x2 x→∞ (7+ 2 x – 3 x2 )′ x→∞ 2 – 6 x

( still

lim

lim x →∞

∞ f orm; apply the ru le) ∞

2x – 3 (2 x – 3)′ 2 1 = lim = lim =– x →∞ x →∞ 2 – 6x (2 – 6 x)′ –6 3

Note L’Hospital’s Rule cannot be applied directly to the indeterminate forms ∞ ⋅ 0 and ∞ – ∞. ∞ 0 But it may be possible to convert them into the form or into the form . ∞ 0

EXAMPLE

8

Solution

f ⋅g=

f g or f ⋅ g = 1 1 g f

1 Find lim x sin . x →∞ x

This limit leads to the form ∞ ⋅ 0, but we can change it to 0 1 the form by writing x = . 1 0 x 1 lim x ⋅ sin = lim x →∞ x x→∞

1 x=0 1 0 x

sin

0 , we can apply L’Hospital’s Rule. 0 1 1 1 1 sin (sin )′ ( − 2 ) ⋅ cos x = lim x = lim x x = lim cos 1 =1 lim x →∞ x→∞ x→∞ x→∞ 1 1 1 x − 2 ( )′ x x x

Since we have the form

Check Yourself 2 Find the following limits: 2 x2 – 5 x +7 x →∞ 3x + 4

1. lim

1 1 ) 5. lim( − x→0 x sin x

Answers 1. ∞ Applications of the Derivative

5. 0 345

EXERCISES

5 .1

0 A. The Indeterminate Form – 0 1. Find the following limits: 2

x – 4x + 4 x2 + x – 6

a. lim x→ 2

1+ 3x – 1 c. lim x →0 x

e. lim x→0

g.

i.

lim

1 – cos x x2

x → –2

lim

sin(2 x + 4) x2 – x – 6 3

3–3x x– 3

x→ 3

x sin 3 b. lim x →0 2 x

2

x – 8x x–2

x→ 2

h. lim x→ 2

j.

3. Find the following limits: 3 a. lim x 3 + x − 2 x→1 2 x − 3 x + 1

› c.

4 x3 + 4 d. lim 2 x → –1 5 x +6 x +1

f. lim

Mixed Problems

3

2 – x+ 6 x2 – 4

x→

2

tan x 1 + tan x

lim(tan x − sec x) x→

π 2

6

d. lim sin x x→0 3x › f.

b. lim π

› e. lim x →1

x −1 πx tan 2 2

1 x2 + 3 x − x ) lim( − csc x) › g. lim( x→∞ x→0 x

2 x2 + 3 x – 2 x →1/ 2 6x – 3 lim

2 k. lim sin x – sin a l. lim tan( x – 4) x→ a x→ 2 x–a 4 – x2 5x x cos + sin arctan 3 x m. lim 6 3 n. lim x →0 arctan 4 x x →π x–π

o. lim x x→8

2/3

–4 x–8

x π arctan( ) – 2 4 p. lim x→ 2 x–2

∞ B. The Indeterminate Form – – ∞

2. Find the following limits: a. lim

x2 + x +1 3 x3 + 4

b. lim

c. lim

x2 + 4 x

d. lim

x →∞

x →∞

9 – x2 x→ – ∞ x – 2 x – 3 x →0

2

cot 3 x cot 5 x

2 e. lim x + cos x x →∞ 3 x2

346

Algebra 10

A. INTERVALS OF INCREASE AND DECREASE In this section, we will first briefly review the increasing and decreasing functions and then discuss the relationship between the sign of the derivative of a function and the increasing and decreasing behavior of the function. Recall the graph of the function f(x) = x2. As we move from left to right along its graph, we see that the graph of f falls for x < 0 and rises for x > 0. The function f is said to be decreasing on (–∞, 0) and increasing on (0, ∞). Definition

y

y=x2

x the graph of f(x) = x

2

increasing and decreasing functions A function f is increasing on an interval I if f(x) increases as x increases on I. That is, for any x1 < x2 on I, f(x1) < f(x2). Similarly, f is decreasing on an interval I if f(x) decreases as x increases on I. That is, for any x1 < x2 in I, f(x1) > f(x2).

a decreasing function

an increasing function

Note We refer to a function as increasing or decreasing only on intervals, not at particular points. Applications of the Derivative

347

We now learn how the first derivative can be used to determine where the function is increasing or decreasing. Let us look at the following graphs. y

a

y

y = f(x)

y = f(x)

x

0° < α < 90° positive slope

a

b x f is increasing positive slopes (f′(x) > 0)

a

b x f is decreasing negative slopes (f′(x) < 0)

Observe that the function f is increasing on the interval (a, b) and the tangent lines to the graph of f have positive slope on that interval. We know that the slope of each tangent line is given by the derivative f ′(x). Thus, f ′(x) must be positive on (a, b). Similarly, we expect to see a decreasing function when f ′(x) is negative. These observations lead to the following important theorem. Theorem

Let f(x) be a differentiable function on the interval I. a. If f ′(x) > 0 for all the values of x on I, then f(x) is increasing on the interval I. b. If f ′(x) < 0 for all the values of x on I, then f(x) is decreasing on the interval I.

Note According to the theorem above, when we are asked to determine the intervals of increase and decrease for a given function, we must examine the sign of the derivative of the function. To do this, we shall construct the sign chart of the first derivative. We assume that you are familiar with constructing the sign chart of a function from your earlier studies. EXAMPLE

9

Solution

Find the intervals where the function f(x) = x2 – 4x + 3 is increasing and where it is decreasing. Let us construct the sign chart of f ′(x).

y

f ′(x) = 2x – 4 and x = 2 is a root of f ′(x) = 0 x –¥

3 2

f ¢(x)

–

f (x)

decreasing ( )

+¥

+ 2

increasing (

)

From the chart, f ′(x) > 0 when x > 2 and f ′(x) < 0 when x < 2. So, f is increasing on (2, ∞) and decreasing on (–∞, 2).

1

3

x

the graph of f(x) = x2 – 4x + 3

It would also be true to say that f is increasing on [2, ∞) and decreasing on (–∞, 2]. 348

Algebra 10

Note As in the previous example, we use an “up arrow” ( ) for the intervals where the function is increasing and a “down arrow” ( ) for the intervals where the function is decreasing.

EXAMPLE

10

Solution

For what values of x is the function f(x) = (x – 1)3 either increasing or decreasing? f ′(x) = 3(x – 1)2.

y

f ′(x) ≥ 0 for any of the values of x because (x – 1) > 0. 2

We conclude that f is always increasing for all the values of x. The graph of f is shown in the figure.

1

x

-1

Note that when x = 1 we have f ′(x) = 0. But that does not affect the increase since it is just a point.

the graph of (fx) = (x – 1)3

EXAMPLE

11

Solution

For what values of a is the function f(x) = ax3 – 2x2 + 2x – 3 increasing for all real numbers? Since f is increasing for all real numbers, f ′(x) > 0. f ′(x) = 3ax2 – 4x + 2 > 0 This is possible only if 3a > 0 and Δ < 0

2

Given ax + bx + c = 0, Δ = b2 – 4ac.

3a > 0. So, a > 0. (1)

2 Δ = (– 4)2 – 4 ⋅ 3a ⋅ 2 = 16 – 24a < 0. So, a > . 3

(2)

2 By (1) and (2), we have a > . 3

12

mx – 2 For what values of m is f ( x) = always decreasing in its x+ 3 domain?

Solution

Since f is always decreasing, f ′(x) < 0 for all the values of x except x = –3.

EXAMPLE

f ′( x) =

m( x + 3) – ( mx – 2) 3 m + 2 = < 0. ( x + 3)2 ( x + 3)2

Then, we have 3m + 2 < 0 because (x + 3)2 is always positive. 2 Thus, m < – . 3 Applications of the Derivative

349

EXAMPLE

13

The graph of the function f is given on the interval (a, b). State whether each of the following functions is increasing or decreasing on (a, b). a. g(x) = x2 – f(x)

b. h(x) = f 2(x)

c.

m( x) =

y y = f (x)

f ( x) x a

Solution

b

x

From the graph, we conclude that x < 0 and f(x) > 0 on (a, b). Also, we have f ′(x) > 0 because f is increasing on (a, b). Now let us find the derivative of each function. a. g′(x) = 2x – f ′(x) < 0. So g is decreasing on (a, b). b. h′(x) = 2 ⋅ f (x) ⋅ f ′(x) > 0. So h is increasing on (a, b). ′ c. m′(x) = f ( x) ⋅ x2− f ( x) < 0. So m is decreasing on (a, b). x

Check Yourself 3 1. Find the intervals where each function is increasing or decreasing. a. f(x) = x3 – 3x2 + 6 2. The function f ( x) =

b. f ( x) =

1 3x + 4

kx +1 is always increasing in its domain. Find k. x +1

Answers 1. a. increasing on (–∞, 0) and (2, ∞), decreasing on (0, 2) 4 b. decreasing on ⎛⎜ −∞, − 4 ⎞⎟ and ⎛⎜ − , ∞ ⎞⎟ 3⎠ ⎝ 3 ⎠ ⎝ 2. k > 1.

B. MAXIMUM AND MINIMUM VALUES 1. Absolute and Local Maximum and Minimum In many applications we need to find the largest or the smallest value of a specified quantity. Here are a few examples: • What is the shape of a container that minimizes the manufacturing costs? • At what temperature does a certain chemical reaction proceed most rapidly? • Which path requires the least time to travel?

These problems can be reduced to finding the maximum or minimum value of a function. Let us first explain what we mean by maximum and minimum values. Definition

absolute maximum and minimum A function f has an absolute maximum at c if f(c) ≥ f(x) for all the values of x in its domain. Similarly, f has an absolute minimum at c if f(c) ≤ f(x) for all the values of x in its domain.

350

Algebra 10

Note Here is some terminology: If a function f has a maximum (or minimum) at x = c, then we say that f(c) is a maximum (or minimum) value of f and (c, f(c)) is a maximum (or minimum) point of f. An extremum of a function is either a maximum or minimum value of that function. The figure on the right shows the graph of a function f with absolute maximum at x = a and absolute minimum at x = d. Note that (a, f(a)) is the highest point on the graph and (d, f(d)) is the lowest point.

y f(a)

y = f(x)

f(d) In the same graph, if we consider only I1 I2 ( ) ( ) the values of x sufficiently near b (for x a b c d example, in the interval I1), then f(b) is the smallest of those values of f(x). In other words, no nearby points on the graph of f are lower than the point (b, f(b)). To define such points, we use the word “local”. So, we say that the function f has a local minimum at the point x = b.

Similarly, f has a local maximum at x = c because f(c) is the largest value of f(x) in the interval I2. We see that no nearby points on the graph are higher than the point (c, f(c)). We now state the formal definition: Definition

local maximum and minimum A function f has a local maximum at c if f(c) ≥ f(x) for all the values of x in an interval I containing c. Similarly, f has a local minimum at c if f(c) ≤ f(x) for all the values of x in an interval I containing c. The figure on the right illustrates some local and absolute extrema of a function f with the domain [a, e]. We see that f has a local maximum at x = c, and a local minimum at x = b and x = d. Also, f has an absolute minimum at x = b and an absolute maximum at x = e. Observe that the absolute minimum is also local, but the absolute maximum is not local because it occurs at the endpoint x = e.

absolute and local max

y local max not absolute

local min not absolute b a

c

d

x

e

y = f(x)

absolute and local min

Applications of the Derivative

351

Note 1. A function has at most one absolute maximum and one absolute minimum. But it may have more than one local maximum or minimum. 2. An absolute extremum of a function is either a local extremum or an endpoint.

Check Yourself 4 1. Explain the difference between an absolute maximum and a local maximum.

y y = g(x)

2. The graph of a function with the domain [a, f] is given on the right. For each of the points from a to f, state whether the function has a local maximum or minimum, or an absolute maximum or minimum. a

Answers

b

c

d

e

f

x

1. absolute max.: f(c) ≥ f(x) for all the values of x in the domain of f. local max.: f(c) ≥ f(x) for all the values of x in an interval I containing c. 2. local max. at x = b, x = d local min. at x = c, x = e absolute max. at x = d absolute min. at x = a.

2. Finding the Local Extrema y We now learn how the first derivative can be used to locate y = f(x) the local extrema. We first consider the functions that have derivatives at the local extremum points. The figure on the right shows the graph of a function f with a local maximum at x = a and a local minimum at x = b. Observe a b that the tangent lines to the graph at these points are x horizontal (parallel to the x-axis) and therefore each has slope 0. Remember that the slope of the tangent line is given by the derivative. So, we say that

f ′(a) = 0, f ′(b) = 0. This analysis reveals an important characteristic of the local extrema of a differentiable function. Theorem

If f has a local extremum at c, and f ′(c) exists, then f ′(c) = 0. 352

Algebra 10

Note

y

y = f(x)

The converse of this theorem is not true in general. That is, when f ′(c) = 0, f does not necessarily have a maximum or minimum at x = c. For example, consider the function f(x) = x3. Here, f′(x) = 3x2, so f′(0) = 0. But, f has neither a local maximum nor a local minimum at x = 0.

EXAMPLE

14

Solution

x

The function f(x) = 2x3 – mx + 5 has a local minimum at x = 1. Find m. Since f(x) is a polynomial function, it is differentiable everywhere. By the theorem above, we have f ′(1) = 0. f ′(x) = 6x2 – m f ′(1) = 0 6 ⋅ 12 – m = 0 m=6

EXAMPLE

15

Solution

In some books, an extremum that occurs at a point where the derivative does not exist is called a singular point.

Find the local extrema of the function f(x) = |x|. Let us plot the graph of f. ⎧⎪ x, x ≥ 0 f ( x) =| x |= ⎨ ⎪⎩− x, x < 0

y

We see that f has a local minimum at x = 0. But there is no horizontal tangent there.

y = f(x)

x

In fact, there is no tangent line at all since f ′(x) is not defined at x = 0.

Note The function f(x) = |x| shows that a local extremum of a function may exist at which the derivative does not exist. As a consequence, we say that the local extrema of any function f occurs at the points c where f ′(c) = 0 or f ′(c) does not exist. Such points are given a special name. Applications of the Derivative

353

Definition

critical point The value c in the domain of f is called a critical point if either 1. f ′(c) = 0, or 2. f ′(c) does not exist.

EXAMPLE

16

Solution

Find the critical points of f(x) = 2x3 – 9x2 + 12x – 7. The derivative of f is f ′(x) = 6x2 – 18x + 12 = 6(x – 1)(x – 2). Since f ′(x) is defined for all the values of x, the only critical points are the roots of f ′(x) = 0. Therefore, x = 1 and x = 2.

EXAMPLE

17

Solution

Find the critical points of f ( x) =

x2 . x –1

The Quotient Rule gives f ′( x) =

2 x( x – 1) – x2 ⋅ 1 x( x – 2) = . ( x – 1)2 ( x – 1)2

Since x = 0 and x = 2 are the roots of f ′(x) = 0, they are critical points. Next, observe that f ′(x) does not exist when x = 1. However, since f is not defined at that point, the point x = 1 is not a critical point.

EXAMPLE

18

In the figure on the right the graph of a function f with critical points at x = a, b, c, d, and e are shown.

y

y = f(x)

a. State why these points are critical. b. Classify each of them as a local maximum, a local minimum, or neither. a

Solution

354

b

c

d

e

x

a. Observe that there are horizontal tangents at the points x = c, d, and e, so f ′(x) = 0 at these points. Next, f ′(x) does not exist at x = a because the tangent line at this point is vertical. Finally, since there is a corner at x = b, f ′(x) does not exist there. Algebra 10

b. From the graph of f, we say that f has a local maximum at x = d, and a local minimum at x = b and x = e. Note that f ′(c) = 0 and f ′(a) does not exist, and f has no local extrema at these points. We conclude that not every critical point gives rise to a local extrema.

Check Yourself 5 Find the critical points of the following functions. 1. f(x) = x3 – 3x + 4

2. f(x) = 1 – ñx

Answers 1. –1, 1 2. 0

3. The First Derivative Test So far we have learned that any extremum of a function f must occur at any critical point of f. In the previous example we have seen that not every critical point is a maximum or a minimum.Therefore, we need a test that helps us classify critical points as local maximum, local minimum, or neither. Suppose that the function f is continuous at c and that f is defined on some open interval containing c. y

If f is increasing on the left of c and decreasing on the right, then f should have a local maximum at x = c.

inc.

dec.

f¢>0 + – f¢<0 + – + – + – c

Applications of the Derivative

x

355

y

dec.

+ – + – + – + –

If f is decreasing on the left of c and increasing on the right, then f should have a local minimum at x = c.

inc.

f¢<0

f¢>0

c y

If f is increasing on both sides or +

decreasing on both sides, then f should have neither a local maximum nor a local minimum at x = c.

y

+

f¢>0

+

+ f¢>0

x

– f¢<0

– – –

+

–

f¢<0

–

+ c

c

x

x

Moreover, we know that f(x) is increasing where f ′(x) > 0 and decreasing where f ′(x) < 0. These observations are the basis of the following test. THE FIRST DERIVATIVE TEST Let c be a critical point of a function f(x). 1. If f ′(x) changes from positive to negative at c, then f(x) has a local maximum at c. 2. If f ′(x) changes from negative to positive at c, then f(x) has a local minimum at c. 3. If f ′(x) does not change sign at c, then f(x) has no local maximum or minimum at c. EXAMPLE

19

Solution

Find the critical points of the function f(x) = x3 – 3x2 – 9x + 6, and classify each critical point as a local maximum, a local minimum, or neither. f ′(x) = 3x2 – 6x – 9 = 3(x – 3)(x + 1) x = 3 and x = –1 are the critical points (where f ′(x) = 0). Since f ′(x) is a polynomial function, it is differentiable everywhere. Thus, we have no points c such that f ′(c) is not defined. x f ¢(x)

–¥

–1

+

3

–

+¥

+

f(x) (max)

(min)

From the sign chart, f increases for x < –1 and decreases for –1 < x < 3. So, f has a local maximum at x = –1. Similarly, f decreases for –1 < x < 3 and increases for x > 3. So, f has a local minimum at x = 3. 356

Algebra 10

EXAMPLE

20

Solution

Find the local extrema of the function f(x) = x2/3 + 2. f ′( x) =

2 –1 2 ⋅ x3 = 3

2 1

3 ⋅ x3 There is no root of f ′(x) = 0.

Now we will look for the values of x such that f ′(x) is not defined but f(x) is defined. We see that f is defined for all the values of x but f ′ is not defined at x = 0. So, 0 is a critical point. x

–¥

f ¢(x)

0

+¥

–

+

f(x) (min)

Thus, the first derivative test tells us that x = 0 is a local minimum of f.

EXAMPLE

21

Solution

Find the local extrema of the function f(x) = |x – 1|. If x > 1, then x – 1 > 0. So, f(x) = x – 1. If x < 1, then x – 1 < 0. So, f(x) = 1 – x. f ′(x) =

1,

x>1

–1,

x<1

Since f ′(1– ) ≠ f ′(1+), f ′(x) is not defined at x = 1. So, x = 1 is a critical point. Furthermore, f ′(x) is not equal to zero anywhere. For x > 1, we have f ′(x) = 1 > 0. So, f(x) is increasing on this interval. For x < 1, we have f ′(x) = –1 < 0. So, f(x) is decreasing on this interval. x

–¥

f ¢(x)

1

–

+¥

+

f(x) (min)

Thus, f has a local minimum at x = 1. Applications of the Derivative

357

EXAMPLE

22

Solution

Find the local extrema of the function f(x) = 5x3 + 4x. f ′(x) = 15x2 + 4 is always positive. There is no real solution of f ′(x) = 0. x

–¥

f ¢(x)

+¥

+

f(x)

f is increasing for all the values of x. Since f(x) is a polynomial function, f(x) is continuous and differentiable everywhere. Thus, f(x) has no local extrema.

Check Yourself 6 Find the local extrema of the following functions. 1. f(x) = 2x2 – 2x + 5

2. f(x) = 1 – x4

Answers 1. min.: x =

EXAMPLE

23

Solution

3. f(x) =

x2 +1 x

4. f(x) = |x2 – x|

1 1 2. max.: x = 0 3. max.: x = –1, min.: x = 1 4. max.: x = , min.: x = 0, x = 1 2 2

If the function f(x) = x3 + ax2 + 15x + b has a local maximum at the point (1, 10), then find a and b. We know that an extremum of a function must occur at a point where f ′(x) = 0 or f ′(x) does not exist. Since f is a polynomial function, f is differentiable everywhere. So, we have f ′(1) = 0. f ′(x) = 3x2 + 2ax + 15 f ′(1) = 0 3 ⋅ 12 + 2a ⋅ 1 + 15 = 0 a = –9 Since the point (1, 10) is on the graph of f, we say that f(1) = 10. f(1) = 10 13 + a ⋅ 12 + 15 ⋅ 1 + b = 10 a + b = –6 b = 3 (since a = –9)

358

Algebra 10

EXAMPLE

24

The graph of the derivative of the function f(x) is shown in the figure. Find the intervals where f(x) is increasing or decreasing and find the local extrema of f.

y

-2

y = f ¢(x)

-1 1

Solution

x

We know that f(x) is increasing when f ′(x) > 0. f ′(x) > 0 means that the graph of f (x) must be above the x-axis. In the figure f ′(x) > 0 for x > 1. So, f(x) is increasing for x > 1. Similarly, f(x) is decreasing when the graph of f ′(x) is below the x-axis. So, f(x) is decreasing for x < 1. x f ¢(x)

–¥

1

–

+¥

+

f(x) (min)

From the chart, f has a local minimum at x = 1. EXAMPLE

25

Solution

For what values of m does the function f(x) = x3 – (m – 1)x2 + 3x – 2 have no local extrema? Since f has no extrema, there must be no root of f ′(x) = 0. f ′(x) = 3x2 – 2(m – 1)x + 3 The equation 3x2 – 2(m – 1)x + 3 = 0 must have no root.

A quadratic equation has 1. two solutions when Δ > 0. 2. one solution when Δ = 0. 3. no solution when Δ < 0.

We need Δ < 0: [–2(m – 1)]2 – 4 ⋅ 3 ⋅ 3 < 0 4(m2 – 2m + 1) – 36 < 0 4(m2 – 2m – 8) < 0 4(m – 4)(m + 2) < 0 Let us construct a chart to solve the above inequality: –2 +

4 –

+

So –2 < m < 4. Applications of the Derivative

359

Check Yourself 7 1. The function f(x) = x3 – 3x2 – 9x + a has a local maximum value of 10. Find a. 2. The graph of the derivative of the function f(x) is given. Find the local extrema of the function f. y

1

–2

2

4

3

–1

x y = f ¢(x)

Answers 1. 5 2. max.: x = 0, x = 4, min.: x = –2, x = 2

4. Finding the Absolute Extrema In most types of maximum-minimum problems, we are more interested in the absolute extrema rather than the local extrema. Recall that the absolute extrema of a function are the largest and the smallest values of that function in its whole domain. The following figures show the graphs of several functions and give the maximum and minimum values of the functions if they exist. y

y

y 1 x

x

–1 x

minimum value is – 1 maximum value is 1

minimum value is 0 no maximum value

no minimum value no maximum value

We have seen that some functions have absolute extrema, whereas other do not. In what conditions does a function have both the absolute maximum and the absolute minimum? The following theorem answers this question. 360

Algebra 10

Theorem

If a function f is continuous on a closed interval [a, b], then f has both an absolute maximum and an absolute minimum on [a, b]. The above theorem guarantees the existence of the absolute extrema of a continuous function on a closed interval [a, b]. Moreover, we know that each absolute extremum can occur either at a critical point in the interior of [a, b] or at an endpoint of the interval. The following steps give a useful method for finding the absolute extrema of a continuous function on [a, b]. CLOSED INTERVAL METHOD 1. 2. 3. 4.

EXAMPLE

26

Solution

Find the critical points of f on the interval [a, b]. Evaluate f(x) at each critical point. Evaluate f(a) and f(b). The largest of the values of f found in Steps 2 and 3 is the absolute maximum, the smallest of these values is the absolute minimum.

Find the absolute extrema of the function f(x) = x2 – 4x + 3 on [0, 3]. Let us apply the Closed Interval Method step by step: 1st Step: To find the critical points of f, we must solve f ′(x) = 0 and also find where f ′(x) does not exist. f ′(x) = 2x – 4 = 0 gives x = 2. The domain of the function is [0, 3]. So, x = 2 is in the domain. And there is no point where f ′(x) is not defined. Thus, the only critical point on [0, 3] is x = 2. 2 Step: f(2) = 2 – 4 ⋅ 2 + 3 = –1 nd

2

y 3

max

3rd Step: f(a) = f(0) = 02 – 4 ⋅ 0 + 3 = 3 f(b) = f(3) = 32 – 4 ⋅ 3 + 3 = 0 4rd Step: fmax[0, 3] = 3, fmin[0, 3] = – 1. The graph of f confirms our results.

Applications of the Derivative

1

2

3 x

-1

y = f (x) min

361

EXAMPLE

27

Solution

Find the maximum and minimum values of the function f(x) = 2x3 + 12x2 + 18x + 6 on the closed interval [–2, 0]. f ′(x) = 6x2 + 24x + 18 = 6(x + 1)(x + 3) f ′(x) = 0 when x = –1 and x = –3. But x = –3 is outside the interval [–2, 0]. So, we do not take it. The only critical point is x = –1. Additionally, we should consider the endpoints of the interval [–2, 0]. Now, we evaluate f(x) at x = –1, –2, and 0: f(–1) = –2 f(–2) = 2 f(0) = 6 fmax[–2, 0] = 6, fmin[–2, 0] = –2.

Notation

fmax[a, b] denotes the maximum value of the function f on the interval [a, b]. fmin[a, b] denotes the minimum value of the function f on the interval [a, b].

EXAMPLE

28

Solution

Find the absolute extrema of the function f(x) = 2 – |x – 1| on [–1, 3]. If x > 1, then x – 1 > 0. So, f(x) = 2 – (x – 1) = 3 – x. If x < 1, then x – 1 < 0. So, f(x) = 2 – (1 – x) = x + 1. ⎧ 3 – x, ⎪ We have f(x) = ⎨ 2, ⎪ ⎩ x + 1, We say that f ′(1) does not

x>1 x=1. x<1 exist because f ′(1–) = 1 and f ′(1+) = –1 are not equal.

So x = 1 is a critical point. Next we consider the endpoints of the interval [–1, 3]. So, we should find the values of f for the points x = 1, –1, and 3. f(1) = 2 f(–1) = 0 f(3) = 0 fmax[–1, 3] = 2, fmin[–1, 3] = 0. 362

Algebra 10

EXAMPLE

29

Solution

x 2 Find the maximum and minimum values of the function f ( x) = + on [–3, 3]. 2 x 1 2 x2 – 4 – 2= 2 x 2 x2 x = –2 and x = 2 are the roots of f ′(x) = 0. So, they are critical points. Next, f ′(x) does not f ′( x) =

exist when x = 0, but f is not defined at this point; so 0 is not a critical point. Now let us find the values of f for critical points and endpoints. 13 13 f (–3) = – , f ( –2) = –2, f (2) = 2, f(3) = 6 6 f max [–3,3]=

13 6

fmin =[–3, 3]= –

13 6

Check Yourself 8 Find the absolute extrema of each function on the given interval. 1. f(x) = x2 – x + 2, [0, 1]

2. g(x) = x3 – 2x2 + x + 1, [–1, 2]

3. h(x) = x – ñx, [1, 4] Answers 1. fmax = 2, fmin =

Applications of the Derivative

7 4

2. fmax = 3, fmin = –3

3. fmax = 2, fmin = 0

363

EXERCISES

5 .2

A. Intervals of Increase and Decrease 1. You are given the graphs of two functions. Determine where the functions are increasing and where they are decreasing. a.

b.

y

6. The function f(x) = ax3 – (a – 2)x2 +

y = f (x)

x

–3 –1

x

7. Find the values of a, so that f ( x) = 2. Find the intervals where each of the following functions is increasing or decreasing. a. f(x) = 3 – 8x

2

c. f(x) = –x + 4x + 3 d. f(x) = x3 + 6x x3 e. f(x) = – 2x2 + 2 3 f. f(x) = 3x4 + 4x3 – 12x2 1 g. f(x) = 2–x h. f(x) = x2/3 5−x i. f(x) = x2 3

x –1 ⎛ ⎝

› k. f ( x) = sin ⎜ x +

always decreasing in its domain.

f(x) = x3 – 3x2 + 3ax + 15 is increasing on (–∞, –2) and (4, ∞), and decreasing on (–2, 4).

9. Let f be an increasing function on (0, ∞). State whether each of the following functions are increasing or decreasing on the same interval. a. –f(x) 1 c. f ( x)

b. x + f(x) d. f(x2)

10. The graph of the function f is π⎞ ⎟ 3⎠

3. Find the intervals where the function f(x) = sin x + cos x is decreasing on [0, 2π].

4. Show that the function f(x) = Arctanx is increasing for all the values of x. 364

x2 − ax is x2 − 4 x + 3

8. Find a, so that the function

b. f(x) = x2 + 1

j. f(x) =

1 x 3

is always increasing for all the values of x. Find a.

2 –1

1 f ( x) = − x3 + mx2 − 4x +1 decreasing 3 for all real numbers?

y y = f (x)

–2

5. For what values of m is the function

given on the closed interval [a, b]. State whether each of the following functions are increasing or decreasing on [a, b]. a. x ⋅ f(x)

y a

b x

y = f(x)

b. f 2(x) + x c. x2 – f(x) Algebra 10

B. Maximum and Minimum Values

15. The function f(x) = x3 – 9x2 + 15x + 7 is given. Find the sum of the x-coordinates of its local extrema.

11. For each of the points from x1 to x9, state whether f has a local maximum or minimum, and an absolute maximum or minimum. y

16. The graph of y = ax2 + bx has an extremum at

y = f (x)

(1, –2). Find the values of a and b.

17. Find k, if f(x) = x3 – 2x2 – 7x + k has a local x1

x2

x3

x4 x5

x6

x7

x8 x9

maximum value of 8.

x

12. Find the critical points of the following functions. a. f(x) = x3 + x

1 x2

b. f(x) = 2x –

c. f(x) = |x + 1|

13.

18. Given that f(x) = x3 + ax2 + bx + 1 has a local maximum at x = –1 and a local minimum at x = 2, find a and b.

y

19. Find the value of m, if the curve y = x3 + 2mx2 + 30 is tangent to the line y = –2.

y = f (x) a

b

c

d

e

x

20. Find the relation between a and b, if the function

In the figure above the graph of a function f with the critical points at a, b, c, d, and e are shown.

f(x) = ax3 + bx + c has one local maximum and one local minimum.

a. State why these points are critical. b. Classify each of them as a local maximum, a local minimum, or neither.

21. For what values of m does the function 2 f(x) = mx + 1 have no local extrema? x −1

14. Find the local extrema of the following functions. a. f(x) = 8x + x2 b. f(x) = –x3 + 3x + 2 x3 c. f(x) = + 2x2 + 4x + 5 3 d. f(x) = (x – 1)2(x + 3)2 x e. f(x) = x +1 16 f. f(x) = x2 – x g. f(x) = |4 – x2| 2

›› h. f(x) = sin x + sin x Applications of the Derivative

22.

y 2 1

3 6

–2 –3

7 x y = f ¢ (x)

The graph of the derivative of a function f is given. Find the local extrema of f. 365

23. Find the absolute extrema of each function on the given interval. 2

28. Given a parametric function y = f(x) with

a. f(x) = 2x – 4x + 3,

[0, 2]

b. f(x) = –x2 + 2x – 1,

[–2, 2]

3

c. f(x) = x – 6x,

[1, 4]

d. f(x) = 2x3 – 15x2 + 24x + 19,

[0, 2]

e. f(x) = x2 – 4ñx,

[0, 3]

f. f(x) = x5 – 5x4 + 1,

[0, 5]

g. f(x) = 9x2 – x4,

[–3, 3]

h. f(x) = 9 – x2 ,

[–1, 2]

i. f(x) = x –

1 , x

[–1, 1]

x–1 k. f(x) = , x +1

[0, 4]

› l.

f(x) = |x – 3x|,

› m. f(x) = 9sin x – sin 3x + 3,

››

[1, 4] [–π, 0]

a. Find the intervals of increase and decrease of f. b. Find the local extrema of f.

29. –3

maximum value of f on [0 , 4].

27. Find the maximum value of f(x) = sinx + cosx. ›

366

2

6

4 5

y = f ¢(x)

The graph of the derivative of a function f is given. a. Find the intervals of increase and decrease of f. b. Find the local extrema of f.

30. At what point does the tangent to the curve ›

y=

x3 – 2 x2 + x – 5 have the smallest slope? 3

31. In the figure, graph of f ′(x)

value of f(x) = x2 – 4x + 8 on [–2, 3].

26. Let f(x) = ax3 – bx. Find a and b, if f(2) = 4 is the

–1

–4

24. Find the sum of the smallest value and the greatest

of f(x) = ax2 + 2x + b, find a + b.

y

›

›

25. If the point (1, 4) is the highest point of the graph

y = 2t2 + 4t + 5 x = t3 + t.

[1, 3]

j. f(x) = 3x2/3,

2

Mixed Problems

is given. Given that the equation f(x) = 0 has only one root and that root is positive, plot a rough graph of f(x).

y y = f ¢(x)

a

b

x

32. Find the range of the function ›

⎧⎪3x4 − 4 x3 − 24 x2 + 48 x, x ≥ 0.5 . f ( x) = ⎨ ⎪⎩8 x3 +12 x2 + 2, x < 0.5

33. For which values of a does the interval ⎡⎢0, 1 ⎤⎥ ›› ⎣ 3⎦

completely include the range of the function 1 f ( x) = 4 ? 3x − 8ax3 +12 a2 x2 + a Algebra 10

A. CONCAVITY In this section we discuss the concept of concavity. As illustrated in the following figures, two increasing graphs on an interval may have different shapes. This depends on how the graphs bend or turn. As we scan the graphs from left to right, we see that the graph of f turns to the left (upward), while the graph of g turns to the right (downward). We say that the function f is concave up on the interval (a, b) and the function g is concave down on the interval (a, b). We now define concavity geometrically. y

y y=f(x)

a

y=g(x)

b

increasing, concave up

Definition

x

a

b

x

increasing, concave down

concavity A function f is concave up on an interval I if the graph of f lies above all of its tangent lines on the interval I. Similarly, f is concave down on I if the graph of f lies below all of its tangent lines on I.

Applications of the Derivative

367

concave up (slopes increasing)

f ′ > 0 ⇔ f is increasing f ′ < 0 ⇔ f is decreasing

concave down (slopes decreasing)

The graphs above illustrates the definition of concavity. Now, we shall see that the second derivative f ′′ tells us where f is concave up and where f is concave down. If f is concave up on (a, b), then the slopes of the tangent lines increase from left to right as shown in the left figure above. This means that the first derivative f′ is increasing on (a, b). We know that if f ′ is an increasing function, then its derivative f ′′ must be positive on (a, b). In a similar way, it can be shown that if f is concave down on (a, b), then f ′′(x) < 0 on (a, b). These observations suggest the following theorem.

Theorem

Let the function f be twice differentiable on the interval I. 1. If f ′′(x) > 0 for all the values of x on the interval I, then f is concave up on I. 2. If f ′′(x) < 0 for all the values of x on the interval I, then f is concave down on I.

EXAMPLE

30

Solution

As a conclusion of the above theorem, we must examine the sign of the second derivative.

Determine where the following functions are concave up and where they are concave down. a. f(x) = 9 – x2

b. f(x) = x3 y

a. f(x) = 9 – x2 f ′(x) = –2x and f ′′(x) = –2 Since f ′′(x) < 0 for all the values of x, f is concave down everywhere. b. f(x) = x3 f ′(x) =3x2 and f ′′(x) = 6x Setting f ′′(x) = 0 gives x = 0.

x f ¢¢(x) f(x)

0

–¥

x +¥

–

+

concave down

concave up

y = 9 – x2

y

y = x3

x

From the sign chart, f′′(x) changes sign from negative to positive at the point x = 0. Observe that the point (0, 0) on the graph of f(x) = x3 is where f changes from concave down to concave up. We call it the inflection point of f. 368

Algebra 10

Inflection point

Definition

An inflection point is a point where a graph changes its direction of concavity. y

y

y

x

a

x

a

f ¢¢(a) = 0

y

a

f ¢¢(a) = 0

x

x

a

f ¢¢(a) does not exist

f ¢¢(a) does not exist

Note At each inflection point, either 1. f ′′(a) = 0 or 2. f ′′(a) does not exist.

FINDING THE INFLECTION POINTS To find the inflection points of a function, follow the steps. 1. Find the points where f ′′(x) = 0 and f ′′(x) does not exist. These points are the possible inflection points of the function f. 2. Construct the sign chart of f ′′(x). If the sign of f ′′(x) changes across the point x = a, then (a, f(a)) is an inflection point of f. EXAMPLE

31

Solution

Investigate f(x) = (x + 1)4 for concavity and find the inflection points. y

f ′(x) = 4(x + 1)3 f ′′(x) = 12(x + 1)2

4

x = –1 is a double root of f ′′(x) = 0. x f ¢¢(x) f(x)

–1

–¥

+¥

+

+

concave up

concave up

x

–1

the graph of f(x) = (x + 1)4

From the sign chart, f(x) is concave up for all the values of x. Also, we have f ′′(x) = 0 when x = –1. But (–1, 0) is not the inflection point of f because f ′′ does not change sign across x = –1. Applications of the Derivative

369

EXAMPLE

32

Solution

Find the intervals of concavity and the inflection points for f ( x) =

1 4 x – 4 x3 +9 x2 – 7 x+5. 2

f ′(x) = 2x3 – 12x2 + 18x – 7 f ′′(x) = 6x2 – 24x + 18 = 6(x – 3)(x – 1) Because f ′′(x) exists everywhere, the possible inflection points are the solutions of the equation f ′′(x) = 0; that is, x = 1 and x = 3. 1

x –¥

3

f ¢¢(x)

+

–

f (x)

concave up

concave down

+¥

+

(inf)

concave up (inf)

From the sign chart for f ′′, we see that f is concave up on (–∞, 1) and (3, ∞) and concave down on (1, 3). It would also be true to say that f is concave up on (–∞, 1] and [3, ∞) and concave down on [1, 3]. Also, observe that f ′′(x) changes sign at x = 1 and x = 3. Therefore, the points (1, f(1)) and (3, f(3)) are the inflection points of f. EXAMPLE

33

Solution

Find the intervals of concavity and the inflection points for f(x) = cos x – sin x on [0, 2π]. f ′(x) = –sin x – cos x and f ′′(x) = –cos x + sin x Since f ′′ is differentiable on [0, 2π], we must find the solutions of f ′′(x) = 0 on [0, 2π]. π 5π f ′′(x) = 0 ⇒ –cos x + sin x = 0 ⇒ tan x = 1 ⇒ x = and x = . 4 4 sin f ¢¢ = 0 +

+

f ¢¢(x)

–

+ – f ¢¢ = 0

–

x

cos

f(x)

5p 4

p 4

0

–

+

concave down (inf)

The sign chart of f ′′ shows that f is concave down on (0, π 5π ( , ). 4 4

2p

–

concave up

concave down (inf)

π 5π ) and ( , 2π) and concave up on 4 4

π π π 2 2 5π 5π 5π 2 2 f ( ) = cos − sin = − = 0 and f ( ) = cos −sin =− + =0 4 4 4 2 2 4 4 4 2 2 π 5π So, ( , 0) and ( , 0) are the inflection points of f. 4 4 370

Algebra 10

EXAMPLE

34

Solution

The function f(x) = x3 + ax2 + bx + 3 has an inflection point at (1, 3). Find the values of a and b. f ′(x) = 3x2 + 2ax + b and f ′′(x) = 6x + 2a We know that the inflection point of f occurs at a point where f ′′(x) = 0 or f ′′(x) does not exist. Since (1, 3) is an inflection point, f ′′(1) = 0. Also we have f(1) = 3 because the point (1, 3) is on the graph of f. f ′′(1) = 0 ⇒ 6 ⋅ 1 + 2a = 0 ⇒ a = –3 f(1) = 3 ⇒ 13 + a ⋅ 12 + b ⋅ 1 + 3 = 3 ⇒ a + b = 0 ⇒ b = 3

EXAMPLE

35

Solution

Find the equation of the tangent line to the curve f ( x) = x2 +

1 at its inflection point. x

We first need to find the inflection point of f. f ′( x) = 2 x −

1 2 and f ′′( x) = 2+ 3 2 x x

Setting f ′′(x) = 0 gives x = –1 and (–1, 0) is the inflection point. f ′′ does not exist when x = 0 but this point is not the inflection point of f. (Why?) Now we can find the equation of the tangent line at the point (–1, 0). The slope of the equation is 1 = −2 − 1 = −3. m = f ′( −1) = 2 ⋅ ( −1) − ( −1)2 Using the point-slope form of a line, y – y1 = m ⋅ (x – x1) y – 0 = –3 ⋅ (x + 1) or y = –3x – 3.

Applications of the Derivative

371

Note In the beginning of this section we have seen that an increasing graph can be either concave up or concave down. This shows that the increase and decrease of a function is independent of the concavity of the function. Remember that the sign of the first derivative determines where f is increasing and decreasing, whereas the sign of the second derivative determines where f is concave up and concave down.

RELATIONSHIP BETWEEN A GRAPH AND ITS DERIVATIVES y

y

y = f(x)

y

y

y = f(x) y = f(x) x

x

y = f(x)

x

x

f ¢ > 0, f increasing

f ¢ > 0, f increasing

f ¢ < 0, f decreasing

f ¢ < 0, f decreasing

f ¢¢ > 0, f concave up

f ¢¢ < 0, f concave down

f ¢¢ > 0, f concave up

f ¢¢ < 0, f concave down

Check Yourself 9 1. Find the intervals of concavity and the inflection points for each function. x +1 x–1 2. The function f(x) = x3 + ax2 + bx + 2 has an inflection point at (1, –1). Find a and b.

a. f(x) = x3 – 2x2 – 7x + 3

b.

f ( x) =

3. The graph of a function y = f(x) is shown in the figure.

y y = f(x)

a. Find the intervals of increase and decrease of f. b. Find the intervals of concavity of f. Answers

1 2 3 4 5 6 7 8 9

x

2 2 2 1. a. concave up: ( , ∞ ) , concave down: (– ∞, ) , inflection point x = . 3 3 3 b. concave up: (1, ∞), concave down: (–∞, 1), no inflection point.

2. a = –3, b = –1 3. a. increasing: (1, 2), (4, 6), (8, 9) b. concave up: (3, 6), (6, 9) decreasing: (2, 4), (6, 8) concave down: (1, 3) 372

Algebra 10

B. THE SECOND DERIVATIVE TEST We have seen that the first derivative test helps us classify the critical points of a function f. Here we learn an alternative test for determining whether a critical point of f is a local maximum or a local minimum.

THE SECOND DERIVATIVE TEST Let f be twice differentiable on an interval I and c be a critical point of f in I such that f ′(c) = 0. 1. If f ′′(c) > 0, then f has a local minimum at x = c. 2. If f ′′(c) < 0, then f has a local maximum at x = c.

y

y y = f(x)

y = f(x) c

c

x

f ¢(c) = 0, f ¢¢(c) > 0

f ¢(c) = 0, f ¢¢(c) < 0

local minimum at c.

local maximum at c.

x

The graphs above illustrates the second derivative test. We know that f is concave up near c if f ′′(c) > 0. This means that the graph of f lies above its horizontal tangent at c and so f has a local minimum at c.

EXAMPLE

36

Solution

Apply the second derivative test to find the local extrema of the function f(x) = x3 – 3x2 – 9x + 6. f ′(x) = 3x2 – 6x – 9 and f ′′(x) = 6x – 6. So, f ′(x) = 0 gives x = –1 and x = 3, the critical points of f. To apply the second derivative test, we compute f ′′ at these points. f ′′(–1) = –12 and f ′′(3) = 12 Since f ′′(–1) < 0, the second derivative test implies that f(–1) = 11 is a local maximum value of f. And since f ′′(3) > 0, it follows that f(3) = –21 is a local minimum value. Remember that we had found the same results by using the first derivative test in Example 24.

Applications of the Derivative

373

Note The second derivative test can be used only when f ′′ exists. Moreover, this test fails when f ′′(c) = 0. In other words, if f ′(c) = 0 = f ′′(c), then there might be a local maximum, a local minimum, or neither at the point x = c. In such cases we must use the first derivative test.

EXAMPLE

37

Solution

Find the local extrema of the function f(x) = 3x5 – 5x3 + 3. f ′(x) = 15x4 – 15x2 = 15x2(x2 – 1) = 15x2(x – 1)(x + 1) = 0 So, the critical points are x = –1, x = 0, and x = 1. The second derivative is f ′′(x) = 60x3 – 30x. When we compute f ′′(x) at each critical point, we find that f ′′(–1) = –30 < 0, f ′′(0) = 0, f ′′(1) = 30 > 0. The second derivative test tells us that f has a local maximum at x = –1, a local minimum at x = 1. Since f ′′(0) = 0, this test gives no information about the critical point 0. Let us apply the first derivative test. x f ¢(x)

–¥

–1

+

0

–

1

–

+¥

+

f(x)

We see that f ′ does not change sign at x = 0. So, f does not have a local maximum or minimum.

Check Yourself 10 Apply the second derivative test to find the local extrema of each function. 1. f (x) = 4x3 + 9x2 – 12x + 7 2. f (x) = 8x5 – 5x4 – 20x3 4 3. f (x) = x + x Answers 1. max.: x = –2 1 min.: x = 2 374

2. max.: x = –1 3 min.: x = 2

3. max.: x = –2 min.: x = 2 Algebra 10

EXERCISES

5 .3

A. Concavity

i. f(x) =

1. The graphs of two functions are given. Find the intervals where the second derivative of the function is positive or negative. a.

y

5– x

f ( x) =

sin x 1 − cos x

4

–2

b.

3

› k. f(x) = –Arcsin(x – 2) ›› l.

2

–4

j. f(x) =

x +1

x

y

3. Find a and b, if f(x) = x4 – 4x3 + ax2 + b has an –2 –3

inflection point at (1, 3). –1

2

x

2. Find the intervals of concavity and the inflection points for each function.

4. The function f(x) = x4 + kx2 + 7x – 7 has an inflection point at x = 1. Find the coordinates of the other inflection point.

a. f(x) = x2 – 5x + 6 b. f(x) = –2x2 + 7x c. f(x) = x3 + x2 d. f(x) = x3 – 3x2 + 5x – 7 e. f(x) = 3x4 – 16x3 + 30x2 + 4 f. f(x) =

in the figure. Find the intervals of concavity and the inflection points of f.

1 4 x – 6x2 + 4x – 7 4

g. f(x) = 3x5 – 10x3 + 5x h. f(x)=

5. The graph of the derivative of a function f is shown

1 x2

Applications of the Derivative

y –1 –3

1

5 3

x y = f ¢(x)

375

B. The Second Derivative Test

9. Show that a cubic function has exactly one inflection point.

6. Determine whether f has a local maximum or minimum at the given value of x, using the second derivative test. 1 a. f(x) = x3 – 3x2 – 7x + 5, x=7 3 b. f(x) = x4 – 3x2 + 2, 1 3

4 3

c. f(x) = 4x + x ,

10. Show that a polynomial function of degree 4 has

x=0

either no inflection point or exactly two inflection points.

x = −1

2 d. f(x) = x2 + , x =1 x

11. Find the equation of the tangent line to the curve y=

7. Find the local extrema of the following functions,

1 3 1 x – 2 x2 + 3x + at its inflection point. 3 3

using either the first or second derivative test. a. f(x) = x3 + 6x2 + 9x + 1 b. f(x) = x4 + 4x3 + 2x2 + 1 x2 c. f(x) = x–2

12. Find a, b, and c so that f(x) = ax3 – 3x2 + bx + c

2

has an inflection point at the point (–1, 1) and a local extremum at x = –2.

d. f(x) = ( x + 3) 3 e. f(x) =

1 1 + x2

f. f(x) = x + sin x

13. The tangent line to the curve f(x) = x3 + 3x2 + cx + 1 at the inflection point of f is perpendicular to the line y = x + 4. Find c.

Mixed Problems 8. Given the graph of f(x), plot a rough graph of

›

f ′(x). y y = f(x)

14. Given that g(x) = e–x ⋅ f(x) where f is a differentiable ›

a

376

b

x

function for all real numbers and the function g(x) has an inflection point at x = a, find an expression for f ′′(a) in terms of f ′(a) and f(a). Algebra 10

In this section we solve applied maximum-minimum problems in which the function is not given directly. When we face such a problem, we are required to first find the appropriate function to be maximized or minimized. The following steps will be helpful for solving these problems. 1. Determine the quantity to be maximized or minimized and label it with a letter (say M for now). 2. Assign letters for other quantities, possibly with the help of a figure. 3. Express M in terms of some of the other variables. 4. Use the data in the problem to write M as a function of one variable x, say M = M(x). 5. Find the domain of the function M(x). 6. Find the maximum (or minimum) value of M(x) with the help of the first derivative. Such problems where we look for the “best” value are called optimization problems.

38

EXAMPLE

Solution

A man has 40 m of fencing that he plans to use to enclose a rectangular garden plot. Find the dimensions of the plot that will maximize the area. We want to maximize the area A of the rectangular plot. Let x and y represent the length and width of the rectangle. Then, since there is 40 m of fencing, 2x + 2y = 40 or x + y = 20.

b a Given a rectangle with sidelength a and b Perimeter = 2(a + b) Area = a ⋅ b

Then we express A in terms of x and y: A = x ⋅ y Expressing A as a function of just one variable, we get A(x) = x ⋅ (20 – x) = 20x – x2 (since y = 20 – x). Since the dimensions will be positive, x > 0 and y = 20 – x > 0 or 0 < x < 20. The derivative is A′(x) = 20 – 2x. So, the only critical point is x = 10. To investigate this critical point, we calculate the second derivative. Since A′′(x) = – 2 < 0, the second derivative test implies that A has a local maximum x = 10. We can verify that this local maximum is the absolute maximum by showing that the graph of A is concave down everywhere. Since A′′(x) < 0 for all the values of x in (0, 20), maximum value of A occurs at x = 10. The corresponding value of y is y = 20 – x = 20 – 10 = 10. Thus, the garden would be of maximum area (100 m2) if it was in the form of a square with sides 10 m.

Applications of the Derivative

377

Note Suppose that f has only one critical point c in the interval I. If f ′′(x) has the same sign at all points of I, then f(c) is an absolute extremum of f on I. This absolute interpretation of the second derivative test is useful in optimization problems.

39

EXAMPLE

Solution

Find two positive numbers x and y such that their sum is 15 and x2 + 5y is as small as possible. We have x + y = 15 and we want to minimize M = x2 + 5y. Expressing M as a function of just one variable we get M(x) = x2 + 5(15 – x) = x2 – 5x + 75 (since y = 15 – x). Since both numbers are positive, x > 0 and y = 9 – x > 0 or 0 < x < 9. 5 The derivative is M′(x) = 2x – 5. So, the critical point is x = . 2 Since M′′(x) = 2 > 0 for all the values of x in (0 , 9), the second derivative test implies that 5 x = is the minimum value of M(x). 2 5 5 25 Therefore, M gets its minimum value when x = , and y = 15 − x = 15 − = . 2 2 2

40

EXAMPLE

Solution

A manufacturer has an order to make cylindrical cans with a volume of 500 cm3. Find the radius of the cans that will minimize the cost of the metal in their production. In order to minimize the cost of the metal, we minimize the total surface area of the cylinder. 500 . πr 2 Hence the surface area of the can as a function of r is

The volume of the can is V = πr2h = 500. So, h = r

A( r ) = 2 πr 2 + 2 πr ⋅ ( h

The surface area of a cylinder is 2πr2 + 2πrh and the volume is πr2h where r is the radius and h the height.

500 1000 ) = 2 πr 2 + , r > 0. πr 2 r

The derivative of A(r) is A′( r ) = 4πr – Thus, the only critical point is r = 3 Since A′′( r ) = 4π +

250 . π

2000 250 > 0 for r = 3 , the second derivative test implies that A gets its 3 π r

minimum value when r = 3 378

1000 4( πr 3 – 250) = . r2 r2

250 . π Algebra 10

EXAMPLE

41

Solution

Find the area of the largest rectangle that has two vertices on the x - axis and another two above the x - axis on the parabola y = 3 – x2. y

Let (x, y) be the vertex of the rectangle in the first quadrant. Then the rectangle has sides with the lengths 2x and y. So, its area is A = 2xy. Using the fact that (x, y) lies on the parabola y = 3 – x2, the expression to be maximized is A(x) = 2 ⋅ x ⋅ (3 – x2) = 6x – 2x3.

(x, y) –ñ3

ñ3 x

The domain of this function is (0, ñ3).

y = 3 – x2

Its derivative is A′(x) = 6 – 6x2. So, the critical points are – 1 and 1. Only the positive value x = 1 lies in the interval (0, ñ3). Since this is the only critical value in the interval, we can apply the second derivative test. The second derivative is A′′(x) = –12x and in particular A′′(x) < 0 for all the values of x in (0, ñ3). So, the maximum value of A(x) in this interval is A(1) = 4. Therefore, the area of the largest rectangle is 4. EXAMPLE

42

Solution

Find the point on the line y = x + 2 that is the closest to the point (1, 2). Let (x, y) be a point on y = x + 2 such that the distance d between (x, y) and (1, 2) is a minimum. 2

We have d = ( x – 1)2 +( y – 2) 2 . Distance between two points (x1, y1) and (x2, y2) is ( x1 – x2 )2 +( y1 – y2 )2 .

If the point (x, y) is on the line, then y = x + 2. To rewrite d in terms of the single variable x, substitute y = x + 2. After substitution, d = ( x – 1)2 + x2 = 2 x2 – 2 x +1.

y=x+2

y

(x, y) d (1, 2)

-2 x

1

It is clear that the minimum of d occurs at the same point as the minimum of d2. So, we minimize d2 to simplify calculations by letting M = d2. M(x) = 2x2 – 2x + 1 has derivative M′(x) = 4x – 2. 1 1 So, M′(x) = 0 when x = . Since M′′(x) = 4 > 0, the minimum value occurs at x = . 2 2 1 5 1 5 Since y = + 2 = , the point ( , ) is the closest point to the point (1, 2). 2 2 2 2 Applications of the Derivative

379

EXAMPLE

43

Solution

The swimmer is 40 m from the shoreline. The lifeguard is 100 m from the point on the shore that is directly opposite the swimmer. The guard can run at a speed of 5 m/s and swim at a speed of 3 m/s. What path should the guard follow to get to the swimmer in the least time? Let x be the distance denoted in the given diagram. swimmer

402 + x2

40 m

guard 100 – x

x

Recall that if travel is at a constant rate of speed, then (distance traveled) = (rate of travel) ⋅ (time elapsed). D In short, D = R ⋅ T or T = . R The total time elapsed is T( x) = (swim time)+( run time) =

swim distance run distance 40 2 + x2 100 – x + = + . swim rate run rate 3 5

We wish to minimize the total time elapsed. So, we differentiate this equation to get T ′( x) =

1 1 1 x 1 −1 ⋅ ⋅ (40 2 + x2 ) 2 ⋅(40 2 + x2 ) ′ − = − . 3 2 5 3 40 2 + x2 5

T′(x) = 0 gives 5x = 40 2 + x2 ⇒ 25x2 = 9(402 + x2) ⇒ 16x2 = 9 ⋅ 402 ⇒ x = ± 30. But x ≠ –30 since x measures a distance. It is left to the student to verify that T′′(30) > 0 which means x = 30 corresponds to a path of minimum time.

Check Yourself 11 1. Find two positive numbers x and y such that their sum is 9 and x2y is as large as possible. 2. A rectangle has area of 144 m2. What dimensions will minimize its perimeter? 3. An open rectangular box with a square base is to be made from 300 cm2 of material. Find the dimensions of the box with the maximum volume. 4. Find the minimum distance from the line 2x + 3y = 13 to the origin. Answers 1. x = 6, y = 3 380

2. 12, 12

3. 5, 10

4. ò13 Algebra 10

How can we explain why a pencil appears to be broken when it is immersed partially into water or why objects under water appear to be nearer the surface than they really are to an observer looking down? It is an illusion caused by the refraction of light. When light passes from one transparent medium to another(like air and water), it changes speed, and bends according to the law of refraction which states: sin θ1 sin θ2 = ν1 ν2 where,

ν1 is the speed of light in medium 1, ν2 is the speed of light in medium 2, θ1 is the angle between the incident ray and normal to the surface at the point P, θ2 is the angle between the refracted ray and the normal.

N

incident ray q1

P q2

medium 1 medium 2 refracted ray

The experimental discovery of this relationship is usually credited to Willebrord Snell (1591-1627) and is therefore known as Snell’s law. Applications of the Derivative

381

Snell’s law can be derived from a physics principle discovered by Pierre de Fermat, the seventeenth-century mathematician. Fermat’s principle states that a ray of light travels the path of minimum time. The derivation of Snell’s law from Fermat’s principle represent an interesting application of the derivative. Suppose a light ray is to travel from A to B, where A is in the medium 1 and B is in the medium 2. Using the geometry of the figure given above, we see that the time it takes the ray to travel from A to B is d d t= 1 + 2 ν1 ν 2 t( x) =

A d1

a

q1 x

2

2

medium 1 medium 2

q2

d2

distance (time= ), velocity

b

d

2

( d − x) + b a +x + . ν1 ν2 2

d–x

We obtain the least time, or the minimum value of t, by taking the derivative of t with respect to x and setting the derivative equal to zero. t′( x) =

x

ν 1 a2 + x2

−

d−x

ν 2 ( d − x)2 + b2

The function t is differentiable for all the values of x. So, the only critical values are the solutions to the equation t′(x) = 0. This equation gives the condition that x

ν 1 a2 + x2 From the figure, we see that sin θ1 = Using these equations, we obtain 382

x a2 + x2

=

d−x

ν 2 ( d − x )2 + b2

and sin θ 2 =

.

d−x ( d − x)2 + b2

sin θ1 sin θ 2 , which is Snell’s law. = ν1 ν2

.

Algebra 10

EXERCISES

5 .4

1. Find two positive numbers such that their sum is 30 and their product is as large as possible.

2. One number is 4 larger than another. How must

7. A man has 120 metres of fencing. He wishes to enclose a rectangular garden adjacent to a long existing wall. He needs no fence along the wall. What are the dimensions of the largest area he can enclose?

they be chosen in order to minimize their product?

8. A closed rectangular 3. The sum of two positive integers is 10. Find the maximum value of the sum of their squares.

4. Find the minimum possible value of the sum of two positive numbers such that their product is m.

box is to be made with 192 cm2 of material. The length of its base is twice its width. What is the largest possible volume of such a box?

y x

2x

5. Find the value of m if the sum of squares of the roots of x2 + (2 – m)x – m – 3 = 0 is to be minimum.

6. A farmer has 100 m of fencing and wants to build

9. An open tank with a square base is to be made of sheet iron. Its capacity is to be 4 m3. Find the dimensions of the tank so that the least amount of sheet iron may be used.

a rectangular pen for his horse. Find the dimensions of the largest area he can enclose.

10. A rectangular sheet of

Applications of the Derivative

x

48 cm

tin, 30 cm by 48 cm, has four equal squares cut out at the corners. The sides are then turned up to form an open rectangular box. Find the largest possible volume of the box.

x

30 cm

383

11. A closed cylindrical drum with the volume 54 m3 is to be manufactured using the minimum amount of metal sheet possible. Find the diameter of the base of the drum.

17. In the figure on the right, ›

12. The sum of two non-negative numbers is 10. Find the minimum possible value of the sum of their cubes.

y

the point (x, y) lies on 4 the graph of the ellipse P(x, y) 2 2 x y + =1 in 9 16 the first quadrant. O A 3 x For what value of y will the area of the triangle POA be maximum?

18. Find the length of the diagonal of the rectangle ›

with the largest area that can be inscribed in an isosceles triangle of base 24 cm and height 10 cm.

13. Find the area of the largest rectangle that is D inscribed in a circle with the radius R. ([AC] is the A diameter of the circle because an angle inscribed in a semicircle is a right angle).

19. A right cylinder is inscribed

C

›

2R B

in a right circular cone with the radius 3 cm and height 5 cm. Find the dimensions of the cylinder of maximum volume.

20. Car A is 125 km directly west of car B and begins 14. What is the shortest distance from a point on the ›

›

curve y2 = 8x to the point (4, 2)?

moving east at 100 km/h. At the same time, car B begins moving north at 50 km/h. At what time t does the minimum distance occur?

15. Find the point on the curve y = lnx that is closest ›

to the line y = x.

16. What is the minimum ›

possible area of a right triangle that is formed in the first quadrant by a line passing through the point (4, 3) and the coordinate axes? 384

y

3

21. Among all tangents to graph of

(4, 3)

4

›› x

45 x +1 at positive x-values, the 4 one which intersects y-axis at maximum y-value y = x 3 − 6 x2 +

is chosen. Find that y-value. Algebra 10

A. ASYMPTOTES When plotting the graph of a function, we need to know the behavior of the function at infinity and the behavior near points where the function is not defined. To describe these situations, we define the term “asymptote”. y = f(x) An asymptote is a line that a curve approaches closer and closer until the distance between the asymptote and the

P

points on the curve approaches zero. x −1 . x+ 2 Observe that f is increasing on the interval (–2, ∞). So, you

Consider the graph of the function f ( x) =

We represent an asymptote by drawing a dashed line.

might think that its values f(x) increase without bound as x increases without bound. (f(x) → ∞ as x → ∞). However, we can see that the graph of f approaches the line y = 1 as x → ∞. So, we say that y = 1 is a horizontal asymptote of f. Next, the function f is not defined at x = –2. Let us examine the behavior of f near –2. We see that the graph of f goes to plus infinity as x approaches –2 from the left, whereas the graph goes to minus infinity as x approaches –2 from the right. So, we say that x = –2 is a vertical asymptote.

Definition

asymptote y y=

x–1 x+1 y=1 x

x = –2

vertical asymptote The line x = a is a vertical asymptote of the graph of f(x) if either lim f(x) = ±∞ or lim f(x) = ±∞. x → a+

x → a–

y

x vertical asymptote

FINDING THE VERTICAL ASYMPTOTE P( x) has a vertical asymptote x = a whenever only the Q( x) denominator of f(x) equals zero (that is, Q(a) = 0 but P(a) ≠ 0).

A rational function f ( x) =

Applications of the Derivative

385

EXAMPLE

44

Solution

Find the vertical asymptotes of the function f ( x) =

x . x –4 2

Let P(x) = x and Q(x) = x2 – 4. Note that x = –2 and x = 2 are the roots of the denominator. Since P(–2) ≠ 0 and P(2) ≠ 0, x = –2 and x = 2 are both vertical asymptotes of the graph of f. horizontal asymptote

Definition

The line y = b is a horizontal asymptote of the graph of f(x) if either

y

horizontal asymptote

lim f(x) = b or lim f(x) = b.

x → –∞

x→∞

x

FINDING THE HORIZONTAL ASYMPTOTE A rational function

f ( x) =

an xn + an −1x n −1 +...+ a1x + a0 bm xm + bm −1x m −1 +...+ b1x + b0

horizontal asympote: lim f ( x) =

x→ ∓ ∞

EXAMPLE

45

Solution

±∞

if

n>m

an / bm

if

n = m.

0

if

n<m

Find the horizontal asymptote of the function f ( x) =

has the following limit as

x . x –4 2

To find the horizontal asymptotes we must evaluate lim f ( x). x →∞

Since the degree of the polynomial in numerator is smaller than the degree of the polynomial x = 0. So, y = 0 is a horizontal asymptote of f. in denominator, lim 2 x →∞ x – 4 EXAMPLE

46

Solution

Find all the vertical and horizontal asymptotes of the function f ( x) =

x2 – x . 2 x – 5x + 3 2

To find the horizontal asymptotes we must evaluate lim f ( x). x →∞

x2 – x 1 = . So, y = 1/2 x →∞ 2 x2 – 5 x + 3 2

Since the degree of x2 – x equals the degree of 2x2 – 5x + 3, lim is a horizontal asymptote of f.

Next, to find the vertical asymptotes we must solve 2x2 – 5x + 3 = 0. 2x2 – 5x + 3 = (x – 1)(2x – 3) = 0 gives x = 1 and x = 3/2. However, x = 1 is also a root of numerator. So, only x = 3/2 is a vertical asymptote of f. 386

Algebra 10

EXAMPLE

47

Solution

EXAMPLE

48

Solution

Find all the vertical and horizontal asymptotes of the function f ( x) =

2 x2 − 3x +5 . x2 +1

2 x2 − 3x +5 = 2. So, y = 2 is a horizontal asymptote of f. x →∞ x2 +1 Since the denominator x2 + 1 is never equal to zero, f has no vertical asymptotes. lim

Find all the asymptotes of the function f ( x) = 2 x3 − 5x2 + 7 x − 12. The function f is a polynomial function. But we can write it as a rational function with 2 x3 − 5x2 + 7x − 12 . 1 Since the denominator is never equal to zero, f has no vertical asymptotes.

the denominator 1 such as f ( x) =

Next, compute lim (2 x3 − 5 x2 + 7 x − 12). x →±∞

We know that the limit of a polynomial at infinity is the limit of the term of highest degree. So, lim (2 x3 − 5 x2 + 7 x − 12) = lim 2 x3 = 2 ⋅( ∞) 3 = ∞. x →+∞

x→+∞

lim (2 x3 − 5 x2 + 7 x − 12) = lim 2 x3 = 2 ⋅( −∞) 3 = −∞.

x →−∞

x→−∞

In other words, lim f ( x) and lim f ( x) do not exist. Therefore, f has no horizontal asymptote. x →+∞

x →−∞

Note A polynomial function has no vertical or horizontal asymptotes. Definition

oblique asymptote The line y = mx + n is an oblique asymptote of the graph of f(x) if either

y oblique asymptotes

lim[ f ( x) − ( mx + n)] = 0 or lim[ f ( x) −( mx + n)] = 0. x →∞

x →−∞

x

To find the equation of an oblique asymptote, we use long division. P( x) For a rational function f ( x) = for which the degree of P is exactly one more than the Q( x) c degree of Q, by dividing Q(x) into P(x), we get f ( x) = mx + n + . Q( x) c In this case, we have lim[ f ( x) − ( mx + n)]= lim = 0. x →±∞ x→±∞ Q( x ) So, the line y = mx + n is an oblique asymptote of the graph of f(x). Applications of the Derivative

387

EXAMPLE

49

Solution

Find all the asymptotes of the graph of f ( x) =

x2 + x . x–2

x = 2 is a vertical asymptote of the graph of f because 2 makes only the denominator zero. Note that the degree of the numerator is one more than the degree of the denominator. So, the graph of f has an oblique asymptote. 6 x2 + x By long division of x – 2 into x2 + x, we can find that f ( x) = = x + 3+ . x–2 x−2 So, y = x + 3 is an oblique asymptote of f.

Check Yourself 12 Find all the asymptotes of the graph of each of the following functions. 1. f ( x) =

x −1 2x + 3

2. f ( x) = −

5x 3 + x2

2. y = 0

3. x =

3. f ( x) =

x2 − 9 −3+7 x − 2 x2

4. f ( x) =

x2 − x − 2 x −1

Answers 3 1 1. x = − , y = 2 2

1 1 ,y=− 2 2

4. x = 1, y = x

B. CURVE PLOTTING Curve plotting is the final part of our study of the derivatives. So far we have seen how to use the derivatives to find the most interesting features of a graph. With the use of all the information about the graph of a function, we can easily draw it. STEPS OF CURVE PLOTTING 1. Domain: Find where f(x) is defined. 2. Intervals of Increase and Decrease: Construct the sign chart of f ′(x) to determine the intervals where f(x) is increasing and where f(x) is decreasing. 3. Local Extrema: Find the critical points of f and classify each as a maximum, a minimum, or neither by using the First Derivative Test. 4. Concavity and Inflection Points: Construct the sign chart of f ′′(x) to determine the intervals where f(x) is concave up and where f(x) is concave down. With the help of the chart, find the inflection points. 5. Intercepts: In y = f(x) setting x = 0 gives the y-intercept and y = 0 gives the x-intercept(s). To find the x-intercept(s) may be difficult, in which case we do not use this information. 6. Behavior at Infinity: Consider lim f(x) and lim f(x) to see how the graph of f x → +∞ x → –∞ behaves as x → ±∞. 7. Asymptotes: Find all the asymptotes of the graph and draw the asymptotes in a coordinate plane by using dashed lines. 8. Graph: Start graphing by plotting the local extrema, inflection points, and intercepts. Then, using the rest of the information, complete the plot by joining the plotted points. 388

Algebra 10

EXAMPLE

50

Solution

Plot the graph of the function f(x) = x3 – 3x – 2.

1. Domain: Recall that the domain of a polynomial function is all real numbers. So, f is defined for all the values of x. 2. Intervals of Increase and Decrease: f ′(x) = 3x2 – 3 = 3(x + 1)(x + 1) When f ′(x) = 0 we have x = –1 and x = 1. –1 x –¥ The sign chart of f ′ shows that f is increasing + – f ¢(x) on (–∞, –1) and (1, ∞) and decreasing on (–1, 1).

1

+¥

+

f(x) (max)

(min)

3. Local Extrema: We have learned that a polynomial function is differentiable everywhere. So, the critical points of f(x) are the roots of f ′(x) = 0. From the results of Step 2, we say that f has a local maximum at x = –1 and a local minimum at x = 1. 4. Concavity and Inflection Points: f ′′(x) = 6x = 0 When f ′′(x) = 0 we have x = 0.

x

The sign chart of f ′′ shows that f is concave down on (–∞, 0) and concave up on (0, ∞).

f ¢¢(x)

So, f has an inflection point at x = 0.

f(x)

–¥

0

+¥

–

+

concave down

concave up

5. Intercepts: x = 0 ⇒ y = –2 (y-intercept) Setting y = 0 leads to a cubic equation. Since the solution is not readily found, we will not use this information. 6. Behavior at Infinity: Recall that the limit of a polynomial function at infinity is the limit of the term of highest degree. lim f(x) = lim (x3 – 3x – 2) = lim x3 = (–∞)3 = –∞

x → –∞

x → –∞

x → –∞

(This means that f(x) decreases without bound as x decreases without bound. So, the graph of f goes to plus infinity as x → –∞) lim f(x) = lim (x3 – 3x – 2) = lim x3 = (+∞)3 = +∞

x → +∞

x → +∞

x → +∞

(This means that f(x) increases without bound as x decreases without bound. So, the graph of f goes to plus infinity as x → +∞) 7. Asymptotes: A polynomial function has no asymptotes. Applications of the Derivative

389

y

8. Graph: We can find f(–1) = 0, f(1) = –4, and f(0) = –2. Plot a local maximum at (–1, 0) a local minimum at (1, –4), an inflection point at (–1, 0), and the y-intercept at y = –2. Finally, complete the graph by passing a smooth curve through the plotted points.

local maximum

f(x) ® +¥ as x ® +¥ 1

–1

x y = f(x)

inflection point

–2 f(x) ® –¥ as x ® –¥

local minimum

–4

It is clear from the graph of f that x = –1 is a root of f(x) = 0. So, x + 1 is a factor of x3 – 3x – 2. The other factor can be found by division: f(x) = (x + 1)(x2 – x – 2) = (x + 1)2(x – 2). Hence x = 2 is also a root of f(x) = 0 and the graph crosses the x-axis at this point. But note that x = –1 is a “double root” of f(x) = 0 and the graph is tangent to the x-axis at x = –1. EXAMPLE

51

Solution

Plot the graph of the function f(x) = –x4 + 8x2 – 7. 1. Domain: Since f is a polynomial, it is defined for all the values of x. 2. Intervals of Increase and Decrease:

x

f ′(x) = –4x3 + 16x = –4x(x2 – 4) When f ′(x) = 0 we have x = –2, x = 0, and x = 2. f is increasing on (–∞, –2) and (0, 2), and decreasing on (–2, 0) and (2, +∞).

–¥

f ¢(x)

–2

0

+

–

2

+¥

+

–

f(x) (max)

(min)

(max)

3. Local Extrema: From the sign chart of f ′(x), f has local maximum at x = –2 and x = 2, a local minimum at x = 0. 4. Concavity and Inflection Points: x

f ′′(x) = –12x2 + 16 = 0

2

When f ′′(x) = 0 we have x = ± f has inflection points at x = −

3

2 3

and x =

2 3

f(x)

.

2 ñ3

2 ñ3

–

+

concave down

concave up

f ¢¢(x)

.

–

–¥

+¥

– concave down

5. Intercepts: x = 0 ⇒ y = –7 (y-intercept) y = 0 ⇒ x1 = –1, x2 = 1, x3 = –ñ7, x4 = ñ7 (x-intercepts) 390

Algebra 10

6. Behavior at Infinity: lim f(x) = lim (–x4) = –(–∞)4 = –∞ and lim f(x) = lim (–x4) = –(+∞)4 = +∞

x → –∞

x → –∞

x → +∞

x → +∞

The graph goes to –∞ to the left and to the right. 7. Asymptotes: Since f is polynomial, it has no asymptotes. 8. Graph: f(–2) = 9, f(0) = –7, f(2) = 9, f ( −

2 3

y

local (and absolute) maximum

5 2 5 , f( )= . 3 3 3

)=

local (and absolute) maximum 9

inflection point (– 2 , 5 ) ñ3 3

( 2 , 5 ) inflection point ñ3 3

5 3 –ñ7

–2 –1

1

2

ñ7

x

y = f(x) –7

f(x) ® –¥ as x ® –¥

f(x) ® –¥ local minimum

as x ® +¥

Check Yourself 13 Plot the graph of each function. 1. f(x) = 2x3 – 3x2 – 12x Answers y 1.

2.

y

y = f(x)

7

–1

2. f(x) = x2(x – 2)2

2

x

y = f(x)

1 1

2

x

–20

Applications of the Derivative

391

EXAMPLE

52

Solution

Plot the graph of the function f ( x) =

x−3 . 2x + 4

1. Domain: Recall that the domain of a rational function is all real numbers except the numbers that make the denominator zero. So, f is defined everywhere except x = –2. 2. Intervals of Increase and Decrease: f ′( x) =

1 ⋅ (2 x + 4) − ( x − 3) ⋅ 2 10 = 2 (2 x + 4) (2 x + 4)2

Since f ′(x) > 0 for all the values of x except –2. So, f is always increasing in its domain. 3. Local Extrema: Note that f ′ does not change its sign. By the first derivative test, we say that f has no local extrema. 4. Concavity and Inflection Points: f ′′( x) =

−40 (2 x + 4)3

The sign chart of f′′ shows that f is concave up on (–∞, –2) and concave down on (–2, –∞). Observe that f ′′ changes its sign at x = –2. But at this point f is not defined. Therefore, there is no inflection point. 5. Intercepts: x = 0 ⇒ y = −

3 ( y - intercept) 4

y = 0 ⇒ x=3

( x - intercept)

x f ¢¢(x) f(x)

–2

–¥

+¥

+

–

concave up

concave down

6. Behavior at Infinity: 1 x−3 lim f(x) = lim = . x → ±∞ x → ± ∞ 2x + 4 2 1 is a horizontal asymptote of the graph of f. 2 Also, x = –2 is a vertical asymptote of the graph of f because –2 makes the denominator zero.

7. Asymptotes: From Step 6, y =

8. Graph:

y y = 1 horizontal asymptote 2

y = f(x) 1 2 –2 x = –2 vertical asymptote

392

–3 4

3

x

Algebra 10

EXAMPLE

53

Solution

Plot the graph of the function f ( x) =

x . x –4 2

1. Domain: The domain of f is all real numbers except x = –2 and x = 2. 2. Intervals of Increase and Decrease: f ′( x) =

1 ⋅ ( x2 – 4) – x ⋅ 2 x – x2 – 4 –( x2 + 4) = 2 = 2 2 2 2 ( x – 4) ( x – 4) ( x – 4) 2

Since f ′(x) < 0 for all the values of x except –2 and 2, f is always decreasing in its domain. 3. Local Extrema: f has no local extrema. 4. Concavity and Inflection Points:

–¥

x

2

2 ⋅ x( x +8) f ′′( x) = ( x2 – 4)3

0

–

f ¢¢(x)

+

concave down

f(x)

f ′′(x) = 0 only when x = 0.

-2

2

–

concave concave up down (inf)

f ′′ is not defined at x = –2 and x = 2.

+¥

+ concave up

Thus, f is concave up on (–2, 0) and (2, ∞) and concave down on (–∞, –2) and (0, 2). The sign of f ′′(x) changes at the points x = –2, 0, and 2. But the only inflection point is x = 0 because f is not defined at –2 and 2. 5. Intercepts: x = 0 ⇒ y = 0 and y = 0 ⇒ x = 0. The point (0, 0) is the only intercept. 6. Behavior at Infinity: lim f(x) = lim

x → ±∞

x → ±∞

x = 0. x –4 2

7. Asymptotes: From Step 6, y = 0 (the x – axis) is a horizontal asymptote of the graph of f. Next, x = –2 and x = 2 are vertical asymptotes of the graph of f because they make the denominator zero. 8. Graph:

y x=–2 vertical asymptote

y = f(x)

–2

x

2

y=0 horizontal asymptote x=2 vertical asymptote

Applications of the Derivative

393

EXAMPLE

54

Solution

Plot the graph of the function f ( x) =

x2 − x + 4 . x –1

1. Domain: f is defined for all the values of x except x = 1. 2. Intervals of Increase and Decrease: f ′( x) =

(2 x − 1) ⋅ ( x − 1) − ( x2 − x + 4) x 2 − 2 x − 3 = ( x – 1)2 ( x − 1)2

When f ′(x) = 0, x = –1 and x = 3. Also, note that f ′(x) is not defined at x = 1.

x

From the sign chart of f ′, f is increasing on (–∞, –1) and (3, ∞) and decreasing on (–1, 1) and (1, 3).

f ¢(x)

–1

–¥

1

+

+¥

3

–

–

+

f(x) (max)

(min)

3. Local Extrema: f has a local maximum at x = –1 and a local minimum at x = 3. 4. Concavity and Inflection Points: f ′′( x) =

x

8 ( x − 1)3

–¥

1

–

+

concave down

concave up

f ¢¢(x)

We conclude that f is concave down on (–∞, –1) and concave up on (1, ∞). But it has no inflection point because –1 is not in the domain of f.

f(x)

+¥

5. Intercepts: x = 0 ⇒ y = –4 (y-intercept) y = 0 ⇒ x2 – x + 4 = 0 ⇒ Δ < 0 (no x – intercepts) 2 6. Behavior at Infinity: lim f(x) = lim x – x + 4 = ±∞. x → ±∞ x → ±∞ x–1

7. Asymptotes: Note that the degree of the numerator of f is exactly one more than the degree of the denominator of f.

y

(3, 5) local min.

So, f has an oblique asymptote. 4 . x–1 So, y = x is an oblique asymptote of f.

By long division, we have f ( x) = x +

5

y=x oblique asymptote

Next, x = 1 is a vertical asymptote of f. –1

8. Graph: f(–1) = –3 and f(3) = 5.

x

3 –3 –4

y = f(x)

394

(–1, –3) local max.

x=1 vertical asymptote

Algebra 10

Check Yourself 14 Plot the graph of each function. x x +1 Answers

1. f ( x) =

2. f ( x) =

2

1.

x +1 2−x

y –1

y = f(x)

1/2 –1/2

2.

1

x

y y = f(x)

–1

1/2

2 x

–1

1

3

5

Applications of the Derivative

2

Fill in the 3 × 3 field of squares such that the graph of the derivative is located below each graph. a

b

c

d

e

f

4

6

395

EXERCISES

5 .5

A. Asymptotes

5. The graph of the function f(x) = a(x – 2)2(x + b) is shown in the figure. Find a and b.

1. Find all the asymptotes of the graph of each function. a. f ( x) =

2 x +1

b. f ( x) =

1 ( x − 1)3

3x + 2 c. f(x) = x – 4x – 5x + 6 d. f ( x) = 4−x 3

y = f(x)

3

2

x

3

2

6. The graph of the function

−3 x e. f ( x) = ( x + 3)2

f. f ( x) = x +1 1 − x2

2 g. f ( x) = 2( x − 1) x + 2x − 3

2 h. f ( x) = 4 −2 x − 3x 2 x + 3x − 9

i. f ( x) =

y

x−3 x2 − 5 x − 6

j. f ( x) =

2

x

–2

7. Find the equation of a

y

polynomial function of degree 4 whose graph is shown in the figure.

4

3x − 1 has exactly one vertical x2 + x + m

asymptote. Find m.

y = f(x)

–1

x3 x2 +9

2 k. f ( x) = x + 2 x + 3 x −1

2. The curve y =

y

f(x) = a(x + b)3(x + c) is shown in the figure. f has an inflection point at x = –1. Find the sum a + b + c.

2

y = f(x)

–1 –2

2

x

8. Plot the graph of each rational function.

B. Curve Plotting

a. f ( x) =

x −1 x +1

b. f ( x) =

c. f ( x) =

x −1 2 x − 2x − 3

2 d. f ( x) = 2( x + 1) x − 4x + 3

3. Plot the graph of each polynomial function. a. f(x) = x3 – 6x2 b. f(x) = (x – 1)2(x + 3) c. f(x) = x3 – 2x2 + x – 2

2 e. f ( x) = x2 − 9 x + 3x

2

d. f(x) = (x – 4)(3 – x)

1 x +1 2

2

f. f ( x) = x + x x−2

e. f(x) = x4 – 2x2 + 1

Mixed Problems

f. f(x) = x(x – 1)(x + 1)2

4. In the figure the graph of a cubic function y = f(x) is given. Find the local minimum value of f.

9. Plot the graph of each function.

y

y = f(x)

a. f ( x) = −x 4 − x2 –1

1 x

–2

396

›

b. f ( x) =

x–2 x+ 2

c. f(x) = sin x + cos2x

10. State how many solutions the equation

› 2x3 + 3x2 – 12x + 3 = a has for each value of a. Algebra 10

CHAPTER REVIEW TEST 1. Find lim x →1

A) 1

5A 5. For what values of k is f(x)=x3 +(k+1)x2 +3x+2

x2 + x – 2 . x2 – 4 x + 3

B) −

3 2

always increasing? D) –

C) 0

3 4

E)

2 3

A) –6 < k < 3

B) k > 0

C) –4 < k < 0

D) –3 < k < 2 E) –4 < k < 2

x 1 – sin 2 2. Find lim . x →π π–x 1 A) 1 B) 2

6. Which of the following is a local extremum of f(x) = x3 – 3x2 + 3x + 1? A) –6 1 D) – 2

C) 0

E) –1

B) –1 D) 2

C) 0 E) no extremum

7. Which of the following is false 3. Find the interval on which f(x) = x2 – 6x + 2 is

for the graph of the function y = f(x)? –2

decreasing. A) (–∞, 3)

C) (3, ∞)

B) (–3, 3)

D) (0, ∞)

E) (–∞, 6)

y

–1

y = f(x) 1

A) f(2) = 0

B) f ′(–1) = 0

C) f ′′(1) > 0

D) f ′′(–1) > 0

2

x

E) f ′(0) < 0

8. Find the interval on which y = (x + 2)3 is 4. Find the value of the limit lim x →1

A) 0

B)

Chapter Review Test 5A

1 2

C) 1

concave up.

3

x –1 . x2 – 1

D)

1 3

E)

1 6

A) (–2, ∞)

B) (–∞, –2)

D) (–∞, ∞)

C) (–2, 2) E) (2, ∞) 397

9. For what value of m does the polynomial

13. Let x1 and x2 be the roots of the equation

P(x) = x4 + x3 + (m – 1)x2 have an inflection point at x = –1?

y = x2 – (m + 1)x + 2m – 1 = 0. Find the value of m that minimizes x12 + x22.

A) –3

A) 0

B) –2

C) –1

10. The graph of the derivative

D) 0

y y = f ¢(x)

of the function f(x) = x3 + ax2 + bx + 1 is given in the figure.

2

Find a + b.

A) 17

B) 11

E) 1

x

–1

C) 5

D) –17

B) 1

C) –1

D) 2

E) –2

14. Find the point on the parabola y =

x2 that is 2

3 closest to the point (– , 0). 2 1 1 A) ( –1, ) B) (1, ) 2 2 1 1 D)(– , ) 2 8

11. Find the intersection point of the asymptotes of

y 5 2

cubic function f, find f(2).

3– x . x+ 2

A) (–2, 3) D) (2, 1)

1 1 E) ( , ) 2 8

E) –10

15. Given the graph of a

y=

C) (0, 0)

1

–1

x y = f(x)

B) (3, –1)

C) (–2, –1)

A)

E) (–1, 3)

3 2

B) 0

C) –1

D) –2

E) –

5 2

16. Which one of the following graphs could be the 12. Which of the following is true for the function f(x) = 2x3 + 3x2 + 12x + 4?

graph of y = x4 – 2x2? A)

B)

y

A) f has a local minimum at x = 0.

x

B) f′(2) < 0

1 C) f is concave up on ( −∞, − ). 2 D) f is always increasing.

E) f has a local maximum at x = –1. 398

D)

y

C)

y

y

x

E)

x y x

x

Algebra 10

5B

CHAPTER REVIEW TEST

1. Let f be an increasing function on the closed

6. If x = a and y = b are the asymptotes of the graph

interval [–4, 4]. Which of the following is definitely true? A) f(3) > 0

B) f(–2) < 0

C) f ′ (1) < 0

D) f(2) < f(–2)

of f ( x) =

x2 + 3x + 2 , find a + b. x2 – 2 x – 3

A) 1

B) 2

B)

2 5

B) 12

C) 3

D) 18

E) 9

2

x +x has a local extremum x+ a

8. Given the graph of the function f, find the slope of the tangent line to the graph of g(x) = x ⋅ f(x) at x = – 3.

at x = 2, find a. 4 5

E) 5

triangle whose hypotenuse is 6 cm. A) 36

A)

D) 4

7. Find the maximum possible area of a right

E) f(–1) < f(1)

2. If the function f ( x) =

C) 3

D) –

C) –1

4 5

E) –

2 5

y 3 –4 –3

x y = f(x)

3. Find the local minimum value of f ( x) = x + 84 .

A) 3

x

A) 1

B)

5 2

C)

3 4

D)

5 4

E)

3 2

B) –4

9.

A)

19 14

B)

9 7

C) 1

D)

17 14

E)

8 7

2

5

4

7 x

Given the graph of the derivative of a function f, what is the sum of the abscissa of inflection points of f? A) –9

5. Which of the following is

E) –3

y = f ¢(x)

–1

–5

4. Find the sum of the maximum and minimum x–2 on the interval [–4, 1]. x–3

D) 0

y –3

values of f ( x) =

C) –6

B) –6

C) 4

D) 7

E) 16

y

false for the graph of the function y = f(x)?

4

–3 –1 –4

2

x

10. Which of the following is

y

the function whose graph is given in the figure?

2

y = f(x)

A) f(x) = (x + 1)2(x – 2) 11 ) <0 5

A) f ′(2) = 0

B) f ′′ (

C) f(–1) = 0

D) f ′(1) > 0 13 E) f ′′ ( – ) < 0 4

Chapter Review Test 5B

B) f(x) = (x + 1)(2 + x)

–1

2

x

y = f(x)

C) f(x) = (x + 1)(2 – x) D) f(x) = (x + 1)2(2 – x) E) f(x) = (x – 1)2(x – 2) 399

11. Which one of the following could be the graph of 2x the function f ( x) = 2 ? x –1 y A) B)

14. Find m if the inflection point of the function 1 3 2 x − x2 + mx + is on the parabola 3 3 y = x2 – 2x + 3. f ( x) =

y

A) 2 –1

C)

–1

1

x

D)

y

1

x

1

x

x

–1

A) –2

1

has a local minimum at the point (1, 2), find a. B) –1

C) 0

D) 1

x2 +1 . Find the distance between x

these points. A) ñ5 400

B) 2ñ5

C) 3ñ5

x2 − 1 has no local extrema? mx + 3

B) –1

C) 0

D) 2

D) 4ñ5

16. Which of the following is

E) 4

E) 5ñ5

y

the function whose graph is given in the figure?

E) 2

13. Let A(x1, y1) and B(x2, y2) be the extremum points of y =

E) –2

x

12. Given that the function f(x) = x4 – ax3 + bx2 – 2x + 3 A) –2

D) –1

15. Which one of following could be the value of m if the function f ( x) =

y

E)

C) 0

y

–1 –1

1

B) 1

–1

1 3

1

3 x

x +1 x2 − 2 x − 3 x +1 C) y = 2 x + 2x + 3

A) y =

E) y =

x −1 x2 − 2 x − 3 x +1 D) y = 2 x +3

B) y =

−1 x2 − 2 x − 3 Algebra 10

Imagine you have a bag of marbles. Your friend asks you how many marbles there are in your bag. If you did not know the answer, you would probably count the marbles: one, two three, four, etc. Matching a word to a marble like this is one way of counting. But we can match other things, too. A prisoner might match every day he spends in his cell to a mark on the wall, or a shepherd might match every sheep he looks after to a pebble in a bag. These are different ways of counting. In these section we will look at different ways of counting.

A. THE ADDITION PRINCIPLE Ali wants to go to the cinema to watch a movie. There are two different halls at the cinema. Three movies are showing in the first hall and four different movies are showing in the second hall. In how many ways can Ali choose a movie to watch? Since Ali cannot be in two different halls at the same time, there are 3 + 4 = 7 different ways for Ali to choose the movie. addition principle

Definition

Let A and B be two actions that cannot both be performed at the same time. If action A can be performed in m ways and action B can be performed in n ways, then the action A or B can be performed in m + n ways.

EXAMPLE

1

Solution

Mary has three different Barbie dolls and two different Cindy dolls. She wants to take them out to play with her friend. However, Mary’s mother will only let Mary choose one doll. In how many ways can Mary choose a doll? Mary has three alternatives for her Barbie dolls and two alternatives for her Cindy dolls. So she can choose a doll in 3 + 2 = 5 ways. In this example, Mary had five different dolls. If we name the dolls D1, D2, ..., D5 then we can list the possible results of Mary’s choice as {D1, D2, ..., D5}. Each element in this set is a possible outcome of Mary’s choice.

402

Algebra 10

B. SYSTEMATIC LISTING In the previous examples it was easy to list the outcomes. Sometimes, however, it can be more difficult: there may be many different outcomes in a problem, or a task may be complicated. In this section we will look at different ways of listing the outcomes of a task or decision.

1. Simple Listing If our task contains only one part, listing the possible outcomes is very easy. For example, if we roll a fair dice there are six different possible results. The list of outcomes is 1, 2, 3, 4, 5, 6.

EXAMPLE

2

Solution

David wants to buy a shirt. There are four different colored shirts in David’s size. In how many different ways can David buy a shirt? Let us assume that the colors of the shirts are blue, red, yellow and green. We can list the available shirts as {blue, red, yellow, green}. Since David only chooses one shirt, each shirt in this set is a possible outcome. So the answer is four.

2. Using a Product Table If our task contains two parts, our listing method is a little different. We can use a table called a product table to list the possible outcomes of the task. Let us look at some examples.

EXAMPLE

3

Solution

Selman needs to go into and out of the library. If the library has two doors, in how many ways can Selman go in and out? There are two tasks for Selman: going into the library and going out. Let the two library doors be A and B. For each task, Selman can choose either door. The product table for Selman’s library visit looks like this: Selman’s library visit Go in

Go out A

B

A

(A, A)

(A, B)

B

(B, A)

(B, B)

We can see that there are four different ways for Selman to go into and out of the library. Probability

403

EXAMPLE

4

Solution

How many two-letter words can be formed from the letters in the set {a, b, c}? This task contains two parts: choosing the first letter and choosing make a product table to list the possible outcomes. Second letter Two-letter combinations a b a aa ab b ba bb First letter c ca cb

the second letter. Let us

c ac bc cc

The table shows us that the outcomes are aa, ab, ac, ba, bb, bc, ca, cb and cc. So there are nine possible words. Of course, these are not real words in English. In problems like this, ‘word’ means a sequence of letters, not a real English word. EXAMPLE

5

Solution

Anton is in his first year at university. He has to take one math or science course in the first term and a different math or science course in the second term. The courses available are Algebra, Geometry, Physics, Biology and Chemistry. In how many ways can Anton choose his two courses? The required task has two parts: choosing the first term’s course and choosing the second term’s course. If we denote each course by its first letter, we can show the possibilities in a table as follows: Anton’s courses

First term

A G P B C

A GA PA BA CA

Second term P B AP AB GP GB PG PB BG BP CG CP CB G AG

C AC GC PC BC

Notice that some pairs are omitted from the table since Anton cannot take the same course twice. So there are 20 distinct possibilities. What would happen if Anton had to take two courses together in the same term? In this case there would only be 10 possibilities. Can you see why?

Check Yourself 1 1. How many two-digit numbers can be formed from the digits 3, 5 and 9? 2. Two dice are rolled and their numbers are added. How many possibilities are there that the result is prime? Answers 1. 9 404

2. 15 Algebra 10

3. Using a Tree Diagram

180

A tree diagram is another useful way to list and count possibilities or outcomes. We use tree diagrams in several subjects. In combinatorics, they help us to cope with some complex problems that cannot be easily solved by using product tables.

30 15 3

6 2 2

3

5 180 = 3 × 5 × 2 × 2 × 3 2 2 =2 ×3 ×5

This tree diagram helps us to find the prime factorization of 180.

EXAMPLE

6

Solution

Joseph has one black and one navy pair of pants. He has also three shirts which are red, yellow and green respectively. In how many different ways can Joseph choose to wear his pants with a shirt? Pants Shirt Result Let us list the choices of pants as {B, N} and the choices of shirts as {R, Y, G}. Since it is not important whether Joseph chooses his pants or shirt first, we may assume that he chooses his pants first. The tree diagram for this problem is opposite. We can see that Joseph has six possible choices.

EXAMPLE

7

Solution

B

R Y G

BR BY BG

N

R Y G

NR NY NG

A student rolls a dice and then tosses a coin. How many different outcomes are possible? This task contains two different parts which occur in an order. Constructing a tree diagram will help us to list the different possible outcomes systematically. Let us list the outcomes of the dice roll as {1, 2, 3, 4, 5, 6} and the outcomes of tossing the coin as {T, H}. The tree diagram shows us that there are 12 different possible outcomes.

Dice 1 2 3 4 5 6

EXAMPLE

8

Solution

Probability

6 results

Coin T H T H T H T H T H T H

Result 1T 1H 2T 2H 3T 3H 4T 4H 5T 5H 6T 6H

In a company, three people applied for a department manager’s position and three different people applied for an accountancy position. Show all the different ways of filling these two positions using a tree diagram. Let A, B and C be the people who applied for the manager’s position. Similarly, let D, E and F be the people who applied for the accountancy position. Then we can construct a tree diagram as follows: 405

Department Accountant manager

Result

A

D E F

AD AE AF

B

D E F

BD BE BF

C

D E F

CD CE CF

We can see that there are nine possible ways to fill the positions.

The data in the tree diagrams we have created so far could also have been shown in a product table. This is because the problems we have looked at contain at most two tasks. However, if we want to use a product table for a question that includes three or more tasks, we will have to construct a three-dimensional table. This is difficult to draw on paper. In this case, a tree diagram is the most appropriate way of listing the outcomes.

Check Yourself 2 Solve each question by making a tree diagram. 1. There are three different routes from city P to city Q and four different routes from city Q to city R. Aydos wants to travel from city P to R through city Q. In how many ways can he do this? 2. How many three-digit numbers can be made from the set {5, 6, 9} if a. any digit can be used more than once in a number? b. each digit can only be used once in a number? 3. An international conference is being held in Merida, Mexico. The Kyrgyz team must choose a flight route from Bishkek to Merida. The possible routes are shown in the following diagram. Houston

Frankfurt Moscow

Merida

Bishkek

Mexico City Barcelona

Istanbul

Regardless of the number of flight connections, how many different possible routes are there? Answers 1. 12 406

2. a. 27

b. 6

3. 7 Algebra 10

C. THE MULTIPLICATION PRINCIPLE In the previous section we studied different ways of listing all the possible outcomes of a particular task. However, writing out the entire list of outcomes may sometimes be very time-consuming and unnecessary. We need a different approach for problems which only ask us to calculate the number of possible outcomes. In many of the problems we have looked at so far, the outcome of each part of a task is equally possible. We say that these tasks satisfy the uniformity criterion. However, in some problems there have been restrictions, for example: two people always have to sit together, or only one person in a team can pilot a certain type of space shuttle. These problems do not satisfy the uniformity criterion. When we need to calculate the number of possible ways of performing a task which satisfies the uniformity criterion, we can use the multiplication principle. multiplication principle (fundamental principle of counting)

Definition

Let a multiple-part task which satisfies the uniformity criterion consist of k parts. If the first part of the task can be performed in n1 ways, the second part can be performed in n2 ways and so on, then the number of ways to perform the entire task is n1 ⋅ n2 ⋅ n3 ⋅.....⋅ nk. EXAMPLE

9

Solution

Nicole has four different skirts, three different blouses and two pairs of shoes which she can wear for a business meeting. In how many ways can Nicole dress for the meeting? We can use a table with three boxes, one for each part of the task. This problem satisfies the uniformity criterion because any of the three blouses can be worn with any of the skirts and shoes. Skirts

Blouses

Shoes

4 choices

3 choices

2 choices

Using n1 = 4, n2 = 3 and n3 = 2, by the multiplication principle Nicole can dress in 4 ⋅ 3 ⋅ 2 = 24 ways. EXAMPLE

10

Solution

How many different two-digit numbers can be formed using the digits in the set {1, 2, 3, 4, 5}? This task has two parts: choosing the tens digit and choosing the units digit. Since there are no restrictions we can start by choosing either the tens digit or the units digit, and each digit can take five different values. Tens

Units

5 choices

5 choices

By the multiplication principle, we can form 5 ⋅ 5 = 25 different numbers. Probability

407

EXAMPLE

11

Solution

In the previous example, how many numbers can be formed if a digit cannot be used twice in a number? This time we cannot use a digit that we have already used. Therefore, we can choose any of the five digits in the set for the tens but only four digits remain for the units. This situation still satisfies the uniformity criterion because we always have five choices for the tens and four choices for the units. Tens

Units

5

4

By the multiplication principle, the number of two-digit numbers that can be formed without using a digit twice is 5 ⋅ 4 = 20.

EXAMPLE

12

Solution

How many three-digit counting numbers are there? It is easy to think that we can use any of the digits 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9 for each place value in a number. The answer therefore seems to be 10 ⋅ 10 ⋅ 10 = 1000, since we can use a digit as many times as we like. However, notice that we cannot put zero in the hundreds place since this would not give us a proper three-digit number. For example, 048 is not a three-digit counting number. Hundreds

Tens

Units

9

10

10

So the number of three-digit counting numbers is 9 ⋅ 10 ⋅ 10 = 900.

EXAMPLE

13

Solution

How many different three-digit odd numbers can be formed using the digits in the set {4, 5, 6, 7, 8, 9}? The number formed must be an odd number. This is a restriction. In such cases we should first consider the digit(s) affected by the restriction. In this question it is the units digit that determines whether the number is odd or even. If the number is odd then the units digit must be 5, 7 or 9. So there are three ways to choose the units digit. We are free to choose the two other digits. Hundreds

Tens

Units

6

6

3

So the answer is 6 ⋅ 6 ⋅ 3 = 108 numbers. 408

Algebra 10

EXAMPLE

14

Solution

In a particular country, the automobile license plates are made up of 2 letters followed by 3 digits. Any of the 26 letters of English alphabet and the digits 0-9 can be used. How many different possible license plates are there? We have 26 possibilities for each letter and 10 possibilities for each digit. So the answer is 26 ⋅ 26 ⋅ 10 ⋅ 10 ⋅ 10 = 262 ⋅ 103 = 676000.

EXAMPLE

15

Solution

A company must form a committee comprising a manager, an assistant and a secretary from a group of nine people. Assuming that any person can do any job, in how many ways can the committee be formed? The restriction here is that no one can hold more than one position. Since the order of selection is not important, suppose that we first select the manager, then the assistant and finally the secretary. Then there are nine possible ways to select the manager, eight ways to select the assistant, and seven ways to select the secretary. Manager

Assistant

Secretary

9

8

7

So there are 9 ⋅ 8 ⋅ 7 = 504 different ways to form the committee.

EXAMPLE

16

Solution

EXAMPLE

17

A briefcase has a five-digit combination lock. The second digit in the combination is 5. At most how many different combinations must we try if we want to open the lock? 10 ⋅ 1 ⋅ 10 ⋅ 10 ⋅ 10 = 104 = 10000

A four-digit number is formed using the digits in the set {1, 2, 3, 4, 5, 6}. a. How many different numbers can be formed? b. How many numbers can be formed if no digit can be used more than once? c. How many numbers greater than 4000 can be formed if no digits are repeated? d. How many numbers less than 3000 and divisible by 5 can be formed if no digits are repeated?

Probability

409

Solution

a. Since there is no restriction and there are six digits, the answer is 6 ⋅ 6 ⋅ 6 ⋅ 6 = 1296. b. Since no digit can be used more than once, the answer is 6 ⋅ 5 ⋅ 4 ⋅ 3 = 360. c. There are two restrictions: the thousands digit must be greater than 3 and no digit can be used more than once. Since our number must be greater than 4000, the possible values for the thousands digit are 4, 5 and 6. Therefore there are  3 ⋅ 5 ⋅ 4 ⋅ 3 = 180 4,5,6 numbers which satisfy the conditions. d. This problem sets three restrictions. First, no digit can be repeated. Second, the number must be divisible by 5, so the units digit must be 5. Third, the number must be less than 3000. So there are two possible numbers for the thousands digit: 1 and 2. Consequently, the total number of possibilities is  2 ⋅ 4 ⋅ 3 ⋅ 

1 =24. 5

1,2

EXAMPLE

18

A three-digit number is formed using the digits {0, 1, 2, 3, 4, 5}. a. How many different numbers can be formed? b. How many different numbers which are greater than 300 and divisible by 5 can be formed if no digit is repeated? c. How many even numbers greater than 200 can be formed if all the digits are different? d. How many numbers divisible by 4 can be formed if all the digits are different?

Solution

a. Provided we do not use zero in the hundreds place, all the other digits can be used without restriction. Hundreds

Tens

Units

5

6

6

So there are 5 ⋅ 6 ⋅ 6 = 180 possible numbers.

b. There are restrictions on the first and last digits. Since the digits cannot be repeated there are two cases: if a number is greater than 500 it will not end with a 5, but any other number can end with zero or 5. First case: Consider the case in which the first digit is 5. Then for the units digit only zero is possible since the number must be divisible by 5. Since 5 and zero are used, there are four possible digits left for the tens place. Hundreds

Tens

Units

1 4 1  

 

 

only 5

1,2,3, or 4

only zero

As we can see, there are four possible numbers (they are 10, 520, 530 and 540). 410

Algebra 10

Second case: Now suppose the number does not begin with 5. Then the hundreds digit will be 3 or 4. There are two possibilities for the units digit: zero and 5. Since two digits have been used for the first and last digits, there are four possible digits left for the tens. Hundreds

Tens

Units

2

4

2

 

 

3 or 4

zero or 5

The answer to the question is the sum of these cases: 4 + (2 ⋅ 4 ⋅ 2) = 20. c. In this question there is a restriction on both the hundreds digit and the units digit. Because we cannot use the digit 2 twice, we need to count the numbers ending in 2 carefully. Let us consider the three possibilities for the units digit. If the units digit is zero:

Hundreds

Tens

Units

4  

4

1  

2, 3, 4, 5

If the units digit is 2:

zero

Hundreds

Tens

Units

3  

4

1  

3, 4, 5

If the units digit is 4:

2

Hundreds

Tens

Units

3  

4

1  

2, 3, 5

4

In conclusion, we can form (4 ⋅ 4 ⋅ 1) + (3 ⋅ 4 ⋅ 1) + (3 ⋅ 4 ⋅ 1) = 40 even numbers greater than 200. d. For a three-digit number to be divisible by 4, the last two digits must be 04, 12, 20, 24, 32, 40, or 52. This means that the last digit in any number we form must be even. If the units digit is zero:

Hundreds 4

Tens

2 1  

 

2 or 4

If the units digit is 2:

Units

zero

Hundreds

Tens

Units

3

3

1

 

 

1, 3 or 5

2

We need to consider the case in which 4 is the units digit in two parts. (Can you see why?) Probability

411

4 in the units place, 2 in the tens place:

Hundreds

Tens

Units

3

1

1

 

 

 

4 in the units place, zero in the tens place:

1, 3, 5

2

4

Hundreds

Tens

Units

4 1 1  

 

 

1, 2, 3 or 5

zero

4

The union of all these possibilities gives us the total number of three-digit numbers which are divisible by 4. So there are (4 ⋅ 2 ⋅ 1) + (3 ⋅ 3 ⋅ 1) + (3 ⋅ 1 ⋅ 1) + (4 ⋅ 1 ⋅ 1) = 24 numbers. Notice that problems b, c and d in Example 26 did not satisfy the uniformity criterion directly. However, we were able to break each problem up into separate cases which satisfied the uniformity criterion and then we added he result of each case. This is a useful strategy when we are solving combinatorics problems.

Check Yourself 3 1. Kamil lives in city E. He wants to go to city G via city F. Four different bus companies travel from E to F and three more bus companies travel from F to G. a. In how many different ways can Kamil travel by bus from E to G? b. Kamil does not want to use the same companies again on his way back home. In how many different ways can he arrange his trip from E to G and back? 2. We have a list of 12 questions for the second part of this chapter. We need to choose one question as an Example, one for a Check Yourself section and one for an Exercises section. Assuming that any question can be used in any section, in how many different ways can we make our choice? 3. An astronomer wants to name 7000 celestial objects with a code made up of two letters from the English alphabet followed by a digit. Is this possible? 4. Almaz’s teacher asks him to write a five-digit number whose first and last digits are even while the others are odd. How many different numbers can Almaz write if he does not want to use the same digit twice? 5. How many three-digit numbers greater than 450 can be formed from the digits in the set {1, 2, 3, 4, 5, 6, 7}? Answers 1. a. 12 b. 72 412

2. 1320

3. no

4. 960

5. 168 Algebra 10

Facial identification is an important part of forensic science. When a person commits a crime, witnesses of the crime can sometimes describe the person’s appearance to the police. A forensic artist works with the witnesses to make a picture of the person’s face, either with pen and paper or with a computer program. This process is called facial reconstruction. Facial reconstruction programs store many pictures (or variations) of different parts of a face: the eyes, nose, ears, mouth and hair, etc. Some of these parts, such as the eyes and nose, are very important in determining the overall appearance of a face. Sometimes the forensic artist guides the witness to keep the different parts of a face consistent and in natural proportion. There are several different facial reconstruction programs. Newer programs allow an operator to construct a face in three dimensions. One program has the following number of variations for the different parts of a face:

The following pictures are examples of some two-dimensional faces produced by the program:

By the multiplication principle, we can calculate that this program can produce over 1030 possible faces using the given variations.

EXERCISES

6 .1

A. The Addition Principle 1. A box contains 5 different white balls and 7 different red balls. In how many ways can we randomly pick a ball from the box?

B. Systematic Listing 2. A house has 5 windows and 2 doors. In how many ways can a burglar break in through a window and get out through a door? Show the possibilities in a product table.

3. How many two-digit prime numbers can we form using the digits 1, 2, 3 and 5?

4. In how many ways we can distribute 3 different gifts to Ömer, Ali and Cihan if each person gets one gift?

5. In how many ways can a president and a secretary be chosen from a group of 4 people if anyone can hold either position?

6. A play-off game is a contest or series of contests that are played to break a tie and determine the winner of a championship. In play-off games, the contests stop as soon as it is clear that one team will win the play-off. A play-off between 2 basketball teams has at most 3 contests. How many different play-offs are possible between the two teams?

7. A family has 3 children. List all the possible gender combinations (male or female) for these children, ordered from oldest to youngest.

8. Sheena needs to go to a school, a restaurant, a cinema and a supermarket. If she must not go to the restaurant before the supermarket, in how many different ways can her trip be arranged? 414

9. A power panel has 5 electric switches and the power supply depends on their positions. The power supply is on unless 2 adjacent switches are both off. How many switch settings will keep the supply on?

10. A group of tourists in a country want to visit 5 cities A, B, C, D and E once. They will fly to each city, beginning at city C and ending at city E. If there is no flight between cities B and D, in how many ways can they organize the journey? Show the possibilities in a tree diagram.

C. The Multiplication Principle 11. A dice is rolled, a coin is flipped and a card is drawn from a deck of 52 cards. How many outcomes are possible?

12. Rashid has 4 different colored pens and wants to color each letter in the word ANTARCTICA so that the same letters are the same color. In how many different ways can he do this?

13. A new scooter is available in 4 different colors with 3 types of engine and 2 types of seat. How many different configurations of this model are possible?

14. n dice are rolled together and all of them show the same number. In how many ways can this be done?

15. An ice-cream shop advertises that it can prepare 126 different varieties of ice cream. An ice-cream variety is determined by the way it is served, its flavour and its topping. It can be served in a bowl, a waffle or a cone and it can be topped with chocolate sauce or ground walnuts or hazelnuts. How many different flavours can be ordered? Algebra 10

16. Igor is preparing a test of 12 multiple-choice

22. A briefcase is locked with two different four-digit

questions for his students. Each question has 4 choices of answer. How many different possible answer keys could Igor prepare?

combination locks. A thief knows that the combination for the first lock is a number from 2000 to 5999 and that the first digit of the second lock is the last digit of the first lock. At most how many different combinations must the thief test in order to open both locks?

17. Dastan is going to take the test described in the previous question. In how many ways can Dastan complete his answer sheet if a. he is not allowed to leave any answer blank? b. he is allowed to leave an answer blank?

23. In how many different ways can we pour tea into 10 cups if a. the cups are identical?

18. Two teachers and 5 students will form a row to have their photos taken. In how many different ways can they be arranged if the teachers must be at the ends of the row?

19. In how many ways can 5 boys and 5 girls be

b. the cups are different?

24. The password on Enzi’s e-mail account is a number with non-repeating digits.The password is at least 2 and at most 4 digits long. How many possibilities are there for Enzi’s password?

seated in a row if the same genders cannot sit next to each other?

25. The auto license numbers of all the cars registered 20. A word which reads the same both forward and backward is called a palindrome. (For example: RADAR is a palindrome.) How many palindromes of 7 letters can be formed using the 26 letters of the English alphabet?

in a particular city are made up of two letters followed by three or four digits. If there are 26 possible letters, how many license numbers can end with 91?

26. An organisation wants to create a registration code 21. The combination for a combination lock has 5 digits. We know that the second and fourth digits are the same and the last digit is odd. How many different possible combinations are there for the lock? Probability

format using the 26 letters A-Z and the digits 0-9. The format for all codes must have 7 characters four digits followed by three letters and there must be enough possibilities to register 200 million different people. Is this possible? 415

27. A man in Italy once suggested the following

35. A machine generates all the possible five-digit

automobile license number format: each number should consist of two letters followed by three digits. If the first digit could not be zero and there are 21 letters in the Italian alphabet, could this format be used to register 400 000 vehicles?

numbers from the digits in the set {0, 1, 2, 3, 4, 5, 6, 7}. How many of them are divisible by 25 if

28. How many two-digit numbers can we form using

a. repetition is allowed? b. repetition is not allowed?

36. How many different numbers with nonrepeating

the elements of the set {1, 2, 3, 4, 5} if

digits from 4000 to 6000 can be produced using the digits {0,1,3,4,5,6,9} if

a. repetition is allowed?

a. the numbers must be odd?

b. repetition is allowed and the number must be even?

b. the numbers must be divisible by 9?

29. In an experiment, a die is rolled 5 times and the numbers obtained are written as digits in order. How many different five-digit odd numbers greater than 40 000 can be produced in this way?

37. The controls on Anton's spaceship are activated with a password. Anton has forgotten the password but he knows that password is a number from 5 digit to 7 digit length which does not begin with zero. At most how many different passwords must Anton try?

30. How many different three-digit odd numbers can be formed using the digits 0, 1, 2, 3, 4 or 5 with no repetition?

38. A security password consists of 7 characters and › each character must be either a digit or a

using the digits 1, 2, 3, 4, 5 or 6 if no two odd digits must be next to each other?

lower-case letter. The password must contain at least one letter and at least one digit. If there are 26 possible letters, how many different passwords can be set?

32. How many numbers between 500 and 1000

39. Bahadýr wants to set a password for his computer.

31. How many three-digit numbers can be formed

inclusive contain repeated digits?

33. How many four-digit numbers have at least one odd digit?

The password must be between 3 and 5 letters long and the first and last letters must be vowels. If there are 26 possible letters, how many different passwords can Bahadýr set?

34. A machine generates all the possible four-digit

40. A monogram is a symbol made up of a person’s

numbers using the digits 1, 2, 3 and 4. In how many numbers is the digit 4 on the left of (but not necessarily next to) 2 if repetition is not allowed?

initials. Explain why in a group of 700 people at least 2 people have a monogram made from the same two-letter monogram.

416

Algebra 10

We can define a permutation as an ordered arrangement of some or all of the elements in a given set. The way a set of books is arranged on a shelf, the seating positions of a group of people at a table or the way the players in a football team line up for a team photo are some examples of permutations since in each case, the order of the elements is important.

A. PERMUTATION FUNCTIONS permutation function

Definition

Let A be a non-empty set. A one-to-one and onto function from A to A is called a permutation function in A. For example, consider the function f: A → A with A = {0, 1, 2, 3} and f(0) = 1, f(1) = 3, f(2) = 0, f(3) = 2. f is shown in the Venn diagram opposite. We can see that it is a one-to-one and onto function, and so it is a permutation function. f = {(0, 1), (1, 3), (2, 0), (3, 2)}.

A

A

.0

.0

.1

.1

.2

.2

.3

.3

⎛ 0 1 2 3⎞ ⎟. Alternatively we can write it in the form f = ⎜ ⎜ 1 3 0 2⎟ ⎝ ⎠ This is a common way of writing a permutation function.

Note that f is not the only permutation which can be defined in A in the example above. In fact, we can define n! different permutation functions in a set with n elements. So in this example we can define 4! = 24 different permutation functions in A. ⎛0 1 2 3⎞ ⎛0 1 2 3 ⎟ , f3 = ⎜ f2 = ⎜ ⎜3 2 0 1⎟ ⎜2 1 0 3 ⎝ ⎠ ⎝ functions.

EXAMPLE

19

Solution

Probability

⎞ ⎛0 1 2 3 ⎟ and f4 = ⎜ ⎟ ⎜0 1 2 3 ⎠ ⎝

⎞ ⎟ are three examples of such permutation ⎟ ⎠

List all the permutation functions defined in K = {p, q, r}. ⎛p q r⎞ ⎛p q r ⎞ ⎛p q r ⎞ ⎛p q r ⎞ ⎛p q r ⎞ ⎟ , f2 = ⎜ ⎟ , f3 = ⎜ ⎟, f 4 = ⎜ ⎟, f5 = ⎜ ⎟, f = f1 = ⎜ ⎜p q r⎟ ⎜p r q⎟ ⎜q p r ⎟ ⎜q r p ⎟ ⎜r p q ⎟ 6 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛p q r ⎞ ⎜ ⎟ ⎜r q p ⎟ ⎝ ⎠ 417

1. Identity Permutation Functions identity permutation function

Definition

Let I be a permutation function defined in a set A. If I(x) = x for every x ∈ A then I is called the identity permutation function in A. For example, if I is the identity function defined in the set P = {1, 2, 3, 4} then I(1) = 1, I(2) = 2, I(3) = 3 and I(4) = 4. ⎛1 2 3 4 ⎞ We can write this identity permutation function as I = ⎜ ⎟. ⎜1 2 3 4 ⎟ ⎝ ⎠

2. Composite Permutation Functions We have already stated that a permutation function in a set A must be a one-to-one and onto function. If f and g are two permutation functions defined in A, then f g and g f are also permutations in A. ⎛1 2 3 4 ⎞ ⎛1 2 3 4 ⎞ ⎟ are two permutation functions defined For example, suppose f = ⎜ ⎟ and g = ⎜ ⎜ 3 2 4 1⎟ ⎜3 1 4 2 ⎟ ⎝ ⎠ ⎝ ⎠ in the set P = {1, 2, 3, 4}. Then the composite function f g is f g(x) = f(g(x)), so f

g(1) = f ( g(1)) = f (3) = 4 ⎫ ⎪ f g(2) = f ( g(2)) = f (2) =1 ⎪⎪ ⎬ , i.e. f f g(3) = f ( g(3)) = f (4) = 2 ⎪ ⎪ f g(4) = f ( g(4)) = f (1) = 3 ⎪⎭

We can visualize this as f o g = f

EXAMPLE

20

Solution

⎛1 2 3 4 ⎞ ⎟. g= ⎜ ⎜4 1 2 3 ⎟ ⎝ ⎠

o

g=

3 1 4 2

o

1 2 3 4 3 2 4 1

=

1 2 3 4 4 1 2 3

.

⎛a b c d⎞ ⎛a b c d⎞ ⎟ are two permutations defined in H = {a, b, c, d}. Show ⎟ and g = ⎜ f =⎜ ⎜b c a d⎟ ⎜d b a c⎟ ⎝ ⎠ ⎝ ⎠ that f g ≠ g f. ⎛a b c d⎞ ⎟ f g=⎜ ⎜ d b a c⎟ ⎝ ⎠

⎛ a b c d ⎞ ⎛a b c d ⎞ ⎜ ⎟ =⎜ ⎟ ⎜ b c a d ⎟ ⎜b a d c ⎟ ⎝ ⎠ ⎝ ⎠

⎛ a b c d ⎞ ⎛ a b c d ⎞ ⎛a b c d ⎞ ⎟ ⎜ ⎟ =⎜ ⎟ g f =⎜ ⎜ b c a d ⎟ ⎜ d b a c ⎟ ⎜d c b d ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 418

1 2 3 4

⎫ ⎪ ⎪⎪ ⎬⇒ f ⎪ ⎪ ⎪⎭

g≠g

f

Algebra 10

Notes

‘∀x ∈ A’ means ‘for all elements x in A’.

1. The composition of permutation functions is not commutative: f

g≠g

f.

2. The composition of permutation functions is associative: ( f ( f g) h = f( g(h(x))) = f ( g h).

h=f

(g

g)

3. For any permutation f and identity permutation I in a set A, f ∀x ∈ A, f

I=I

h) because

f = f since

f(x) = I(f(x)) = f(x).

I(x) = f(I(x)) = f(x) and I

3. The Inverse of a Permutation Function Since a permutation f in a set A is both one-to-one and onto, by reversing the ordered pairs of f we get the inverse permutation function of f, denoted by f –1. For example, if P = {0, 1, 2, 3} is a set and ⎛0 1 2 3 f =⎜ ⎜3 1 0 2 ⎝

Note that f EXAMPLE

21

Solution

22

–1

= f –1

⎞ ⎟. ⎟ ⎠

f = I.

⎛Δ †  œ  ⎞ ⎛Δ †  œ ⎞ ⎟ then g−1 = ⎜ ⎟. If g = ⎜ ⎜†  Δ œ ⎟ ⎜ Δ  œ †⎟ ⎝ ⎠ ⎝ ⎠ ⎛ Δ † œ  ⎞ ⎛ Δ †  œ  ⎞ ⎛ Δ †  œ  ⎞ ⎟ ⎜ ⎟= ⎜ ⎟ = I. g −1 = ⎜ ⎜ †  Δ œ ⎟ ⎜ Δ  œ † ⎟ ⎜ Δ †  œ  ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

⎛1 3 5 7 9 ⎞ ⎟ and g are two permutations defined in the set K = {1, 3, 5, 7, 9}. f =⎜ ⎜ 37 9 5 1 ⎟ ⎝ ⎠ g

Solution

f

⎞ ⎛0 1 2 3 ⎟ , i.e. f −1 = ⎜ ⎟ ⎜2 1 3 0 ⎠ ⎝

⎛Δ †  œ  ⎞ In a set K = {{,‹,œ,…,U}, the permutation function g = ⎜ ⎟ is defined. ⎜†  Δ œ ⎟ ⎝ ⎠ –1 Show that g g = I.

So g

EXAMPLE

⎛3 1 0 2 ⎞ −1 ⎟ is a permutation in P then f = ⎜ ⎜0 1 2 3 ⎟ ⎝ ⎠

⎛1 3 5 7 9 ⎞ ⎟ is given. Find g. f =⎜ ⎜ 75 3 9 1 ⎟ ⎝ ⎠

To find g, we have to eliminate f from g We can achieve this by composing g (g

f)

–1

f =g

(f

–1

f )=g

f.

f with the inverse of f, since

I = g.

⎛1 3 5 7 9 ⎞ ⎛1 3 5 7 9 ⎞ ⎟ then f −1 = ⎜ So we must find f –1. If f = ⎜ ⎟. ⎜ 37 9 5 1 ⎟ ⎜9 1 7 3 5⎟ ⎝ ⎠ ⎝ ⎠

So g = (g

f)

f

–1

⎛1 3 5 7 9 ⎞ ⎟ =⎜ ⎜ 75 3 9 1 ⎟ ⎝ ⎠

⎛1 3 5 7 9 ⎞ ⎛1 3 5 7 9 ⎞ ⎜ ⎟= ⎜ ⎟. ⎜ 9 1 7 3 5 ⎟ ⎜1 7 9 5 3 ⎟ ⎝ ⎠ ⎝ ⎠

given

Probability

419

Check Yourself 4 ⎛0 1 2 3 1. f = ⎜ ⎜3 2 1 0 ⎝

⎞ –1 ⎟ is defined in K = {0, 1, 2, 3}. Find f . ⎟ ⎠

⎛2 4 6 8 2. g = ⎜ ⎜6 2 8 4 ⎝

⎞ ⎟ is defined in P = {2, 4, 6, 8}. Find g ⎟ ⎠

g.

⎛ Δ †  œ⎞ 3. g = ⎜ ⎟ and f are defined in H = {€, œ, …, Δ}. g ⎜† œ  Δ ⎟ ⎝ ⎠

⎛ Δ †  œ⎞ ⎟ is given. Find f . f =⎜ ⎜†  Δ œ ⎟ ⎝ ⎠

Answers ⎛0 1 2 3 1. ⎜ ⎜3 2 1 0 ⎝

⎞ ⎟ ⎟ ⎠

⎛2 4 6 8 ⎞ 2. ⎜ ⎟ ⎝8 6 4 2 ⎠

⎛Δ †œ⎞ 3. ⎜ ⎟ ⎝Δ œ † ⎠

B. PERMUTATIONS OF n ELEMENTS We have defined a permutation as ordered arrangement of a set of elements or items. In a permutation, the order of the items is important. We considered some permutation problems in our study of the multiplication principle. Here is another example of a permutation problem: in how many different ways can the three students Faruk, Oleg and Evgeny be seated at a desk?

420

Algebra 10

permutation

Definition

An ordered arrangement of some or all of the elements of a given set is called a permutation. The number of permutations of all of the n distinct elements in a set is denoted by P(n, n), where P(n, n) = n(n – 1) ⋅ (n – 2) ⋅...⋅ 2 ⋅ 1 = n!. In our library seating problem we can see that there are six ways for three students to sit at a desk. Using the permutation notation described above, we can write P(3, 3) = 6.

EXAMPLE

23

Solution

What is the number of permutations of 5 different math books piled on a table? By the definition above, the answer is P(5, 5) = 5! = 120 permutations because there are five distinct books. We can check this answer using the counting technique we learned when we studied the multiplication principle: First book 5

Second book Third book Fourth book 4

3

2

Fifth book 1

Again we find that there are 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1 = 120 ways to put the books in a pile.

EXAMPLE

24

Solution

EXAMPLE

25

Solution

Probability

How many different eight-letter permutations are there of the letters in the word ISTANBUL? Notice that there are eight letters and each letter is distinct. So we can use the formula for P(n, n) using n = 8: P(8, 8) = 8! = 40320 permutations.

A football league has 18 teams. How many different rankings from first to last are possible in the end-of-season league table, assuming that there are no ties? Since each football team is distinct we can use the permutation formula. Therefore the answer is P(18, 18) = 18! possible rankings. 421

EXAMPLE

26

Murat has 5 different math books, 3 different biology books and 4 different physics books. In how many different ways can Murat arrange his books a. on a book shelf? b. in three different file holders, if each holder is for a different subject?

Solution

a. There is no restriction on the order of the books on the shelf so we do not need to consider the subjects. Since there are twelve books, the answer is P(12, 12) = 12! different ways. b. In this case we need to consider the subjects separately. P(5, 5) ⋅ P(3, 3) ⋅ P(4, 4) ⋅ 3! = 5! ⋅ 3! ⋅ 4! ⋅ 3! = 120 ⋅ 6 ⋅ 24 ⋅ 6 = 103 680 arrangements. math

biology

physics

Check Yourself 5 1. In how many different ways can 5 students form a queue? 2. Rasim, Togrul and Elnur are going to establish a company whose name will be a combination of their initials. How many company names are possible? 3. There are 10 desks in a classroom and each desk has two seats. In how many different ways can 20 students sit in the classroom? Answers 1. 120

2. 6

3. 20!

These toys are some interesting samples for permutation puzzles.

422

Algebra 10

C. PERMUTATIONS OF r ELEMENTS SELECTED FROM n ELEMENTS Many permutation problems ask us to consider arrangements of r things chosen from n things (0 ≤ r ≤ n), i.e. permutations of r elements chosen from a set of n elements.

EXAMPLE

27

Solution

How many different two-letter combinations can we form from the letters of the word KANO if a letter cannot be used more than once? The order of the letters is important and a letter cannot be used more than once. By the multiplication principle, the number of combinations is: 4 ⋅ 3 = 12. These combinations are KA KN KO

AK AN AO

NK NA NO

OK OA . ON

In this section we will use a new formula to solve problems of this type.

permutation of r elements selected from n elements

Definition

The number of permutations of r elements selected from a set of n elements is n! P( n, r ) = ( n, r ∈ and 0 ≤ r ≤ n). ( n − r )! If we apply this formula to Example 39, we can write the answer as Some books use nPr or Pnr to mean P(n, r).

P(4, 2) =

4! 4! = =12. (4 − 2)! 2!

Note Any question which can be solved using this permutation formula can also be solved using the multiplication principle.

EXAMPLE

28

Solution Probability

Calculate P(5, 3) ⋅ P(7, 2). P(5, 3) ⋅ P(7, 2) =

5! 7! 5! 7! ⋅ = ⋅ = 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 = 2520 (5 − 3)! (7 − 2)! 2! 5! 423

EXAMPLE

29

Solution

EXAMPLE

30

Solution

Evaluate the expressions. a. P(7, 3)

b. P(n, n)

c. P(n, 0)

a. P(7,3) =

7! 7! 7 ⋅ 6 ⋅ 5 ⋅ 4! = 210 = = (7 − 3)! 4! 4!

b. P( n, n) =

n! n! n! = = = n! ( n − n )! 0! 1

c. P( n,0) =

n! n! n! = = =1 ( n − 0)! n! n!

P(n, 3) ⋅ 5 = P(n, 4) is given. Find n. P(n, 3) ⋅ 5 = P(n, 4) n! n! ⋅5= ( n − 4)! ( n − 3)! 5 1 = ( n − 3) ⋅ ( n − 4)! ( n − 4)! 5= n − 3 n=8

EXAMPLE

31

Solution

How many three-digit numbers can be formed from the digits in the number 13567 if a digit cannot be used more than once? We are choosing three digits from five digits. So there are P(5, 3) =

5! 5 ⋅ 4 ⋅ 3 ⋅ 2! = 60 different three-digit numbers. = (5 − 3)! 2!

Notice that we could have solved the same question using the multiplication principle: 5 ⋅ 4 ⋅ 3 = 60. EXAMPLE

32

Solution

Three raffle tickets will be selected in order from a box containing 30 tickets. The person holding the first ticket will win a car, the person with the second ticket will win a computer, and the person with the third ticket will win a CD player. In how many different ways can these prizes be awarded? Since the question is about an ordered arrangement of three tickets selected from thirty tickets, we can use the formula: P(30, 3) =

424

30! 30! 30 ⋅ 29 ⋅ 28 ⋅ 27! = = = 24360. (30 − 3)! 27! 27! Algebra 10

Remark

The number of permutations of r elements selected from n elements is the product of r successive numbers less than or equal to n: P( n, r ) = n ⋅ ( n − 1) ⋅ ( n − 2) ⋅... ⋅( n − r +1). r factors

For example, P(5, 4) = 5 ⋅ 4 ⋅ 3 ⋅ 2 = 120, P(10, 3) = 10 ⋅ 9 ⋅ 8 = 720 and P(20, 1) = 3 factors

4 factors

EXAMPLE

33

Solution

20

= 20.

1 factor

A fighter plane has seats for a pilot and a copilot. In how many different ways can these be selected from a squadron of 18 soldiers? P(18, 2) =18 ⋅17 = 306 2 factors

EXAMPLE

34

Solution

Mutually exclusive cases are cases which cannot happen at the same time.

How many different combinations of at least 3 letters can be formed from the letters in the word MATHS if no letter can be used more than once? ‘At least 3 letters’ means the combination can have 3 letters, 4 letters or 5 letters. So we need to consider three mutually exclusive cases: combinations of 3 letters, 4 letters and 5 letters. Then we add the number of permutations in each case: P(5, 3) + P(5, 4) + P(5, 5) = (5 ⋅ 4 ⋅ 3) + (5 ⋅ 4 ⋅ 3 ⋅ 2) + (5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 1) 3 - letter words

4 - letter words

5 - letter words

= 60 + 120 + 120 = 300.

So there are 300 possible combinations.

EXAMPLE

35

Solution

Kemal’s bookcase has three shelves. Kemal has 5 different math books, 6 different biology books and 7 different physics books. He wants to arrange 3 math books, 4 biology books and 5 physics books on the shelves so that each shelf is for one subject only. In how many different ways can Kemal arrange his books? There are P(5, 3) possible ways to arrange the math books. There are also P(6, 4) and P(7, 5) different possible ways to order the biology and physics books respectively. However, Kemal can choose the shelves for the subjects in 3! ways. As a result there are P(5, 3) ⋅ P(6, 4) ⋅ P(7, 5) ⋅ 3! = 60 ⋅ 360 ⋅ 840 ⋅ 6 math

biology

physics

shelves

= 108 864 000 ways for Kemal to arrange his books. Probability

425

EXAMPLE

36

A three-digit number is formed by choosing elements from the set {1, 3, 4, 5, 7, 8, 9} without repetition. a. How many numbers do not contain the digit 5? b. How many numbers contain the digit 5? c. How many numbers contain 1 or 7 or both 1 and 7?

Solution

a. The problem is the same as finding the number of three-digit permutations of the set {1, 3, 4, 7, 8, 9} (5 excluded): P(6, 3) = 6 ⋅ 5 ⋅ 4 = 120. b. The total number of three-digit permutations of the set {1, 3, 4, 5, 7, 8, 9} is P(7, 3) = 7 ⋅ 6 ⋅ 5 = 210. From part a, 120 of these permutations do not contain the digit 5. So there are 210 – 120 = 90 three-digit numbers which contain the digit 5. c. We begin by calculating the number of three-digit permutations in which 1 and 7 are not used: P(5, 3) = 5 ⋅ 4 ⋅ 3 = 60 permutations. So there are 210 – 60 = 150 three-digit numbers which contain 1 or 7 or both 1 and 7.

Check Yourself 6 1. There are 7 different pieces of fruit on a tray. We will choose 3 of them and arrange them in a row on a plate. How many different arrangements are possible? 2. The students in a class are photographed in pairs such that each student is photographed with every other student. If there are 90 photos, how many students are there in the class? 3. A machine generates all the possible two-letter combinations of the letters ABCDE, without using a letter twice. What percentage of the combinations do not contain a consonant? 4. How many of the four-digit numbers formed from the digits of the number 12345 without repetition do not begin with the digit 2? 5. A group A contains 6 students and a group B contains 8 students. In a class photo, two students who are to sit in the front will be from A and three students who are to stand at the back will be from B. How many arrangements are possible? Answers 1. P(7, 3) = 210 426

2. 10

3. 10%

4. P(5, 4) – P(4, 3) = 96

5. P(6, 2) ⋅ P(8, 3) = 10080 Algebra 10

D. PERMUTATIONS WITH RESTRICTIONS 1. Permutations with Grouped Elements Sometimes a permutation problem states that we should not separate two or more elements in the set. In this case we count the elements as a single element. Then we apply the ordinary permutation rules. However, we need to consider the number of arrangements within the group of combined elements. By the multiplication principle, we multiply the results to get the answer.

EXAMPLE

37

Solution

How many five-letter words can we form using all the letters in the word MERAK if A and K must be next to each other? First we count the group of letters A and K as a single element. Then the problem is to find the number of permutations of four elements, namely M, E, R and (A, K). However, for each permutation of these four elements there are two arrangements within the group (A, K). Therefore the answer is (

3

+

M , E, R

EXAMPLE

38

Solution

1

) ! ⋅ 2!

group of A and K

= 4! ⋅2! = 24 ⋅2 = 48 words.

A, K

Solve the roller-coaster problem in Example 11 by using grouping and permutation. Since Ahmet and Cemal must sit together, we count them as single element. (

2

+

Berk and Deniz

1 group of Ahmet and Cemal

)! ⋅

2!

= 3! ⋅2!

Ahmet , Cemal

=12. This is the answer we found in Example 11.

EXAMPLE

39

Solution

In how many ways can the children Anar, Maksat, Sasha, Dilshat, Catalin and Mehmet sit in a row if Maksat and Catalin must not sit together? The total number of possible arrangements of six children is 6! = 720. From these, the number of permutations in which Maksat and Catalin sit together will be (4 + 1)! ⋅

2!

= 5! ⋅ 2! = 240.

Maksat , Catalin

So the number of permutations in which Maksat and Catalin are not together is 720 – 240 = 480. Probability

427

EXAMPLE

40

In Example 38, Murat had five different math books, three different biology books and four different physics books. In how many ways can Murat arrange his books on a shelf if a. the math books must be kept together? b. the biology and physics books must be kept together in two different groups? c. all the books on the same subject must be kept together?

Solution

a. Since the math books must be together we consider them as a single book. So the answer is (

1

+

group of math books

b. (

+

1

math books

+

1 group of biology books

EXAMPLE

41

Solution

+

4

+

group of math books

⋅

3! biology books

1 )! ⋅ group of physics books

= 8! ⋅ 5!.

math books

1 )! ⋅ group of physics books

1

)! ⋅ 5 !

physics books

+

5

group of biology books

c. (

3 biology books

= 7! ⋅ 3! ⋅ 4!

4! physics books

3!

⋅

biology books

4!

⋅ 5!

physics books

= 3! ⋅ 3! ⋅ 4! ⋅ 5!

math books

In the My Documents folder on my computer there are 3 files with the extension jpg, 5 files with the extension txt, and 6 files with the extension hzm. In how many ways can the files be listed if the files are ordered by type, ignoring alphabetical order?

⎛ ⎜ ⎝

1

+

group of hzm files

+

1 group of jpg files

1 group of txt files

⎞ ⎟! ⋅ ⎠

6!

⋅

hzm files

3! jpg files

⋅

5!

= 3! ⋅ 6! ⋅ 3! ⋅ 5!

txt files

Check Yourself 7 1. A company owns 3 different green cars, 2 different red cars, one blue car and one yellow car. In how many different ways can they be parked in the company’s parking lot so that cars of the same color are parked together? 2. Five different countries each send 3 people to an international meeting. A photographer wants to photograph all the people in a row such that people from the same country stand together. How many different photographs are possible? Answers 1. 4! ⋅ 3! ⋅ 2! 428

2. 5! ⋅ 3! ⋅ 3! ⋅ 3! ⋅ 3! ⋅ 3! Algebra 10

2. Permutations with Identical Elements Remember that order is important in a permutation: for three objects A, B and C, the permutations (A, C, B) and (B, A, C) are different. Now suppose that we are asked to find the number of permutations of the letters A, A, B, C, D. The number of permutations of five objects is P(5, 5) = 5! = 120. However, some of these of these permutations will be the same because there are two A’s in the list. For example, let A1 and A2 be the two A’s. Then the permutations (A1, B, A2, D, C) and (A2, B, A1, D, C) are indistinguishable. In order to find the number of distinguishable permutations we can use the following formula:

distinguishable permutations of a set with identical elements

Theorem

Let A be a set of n elements which has n1 of one kind of element, n2 of a second kind, n3 of a third kind, and so on, where n1 + n2 + n3 + ... +nr = n . Then the number of distinguishable permutations in A is

EXAMPLE

42

Solution

n! . n1 ! ⋅ n2 ! ⋅ n3 ! ⋅ ... ⋅ nr −1! ⋅ nr !

What is the number of permutations of the digits in the number 5711? By the formula above, there are

4! = 12 different permutations. 2! ⋅ 1! ⋅ 1!

To check our answer let us list the possible permutations. To avoid confusion between the two 1 digits we will name them 11 and 12. 571112 511712 511127 751112 711512 711125 115712 115127 117512 117125 111257 111275 571211 512711 512117 751211 712511 712115 125711 125117 127511 127115 121157 121175 We can see that the permutations in the bottom row are the same as those in the top row. In the bottom row, we have simply swapped the positions of the two identical digits. So there are indeed 12 different permutations. EXAMPLE

43

Solution

How many distinguishable permutations are there of the letters in the word NAHCIVAN? The letters A and N each occur twice. By the formula, the answer is

Probability

8! 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2! = 8 ⋅ 7 ⋅ 6 ⋅ 5 ⋅ 3 ⋅ 2 = 10080. = 2!⋅ 2! 2! ⋅ 2 ⋅ 1 429

EXAMPLE

44

Solution

EXAMPLE

45

Solution

In a kitchen there are 3 identical porcelain dishes, 2 identical metal dishes and 4 identical glass dishes. In how many ways can they be piled up? n!total 9! = = 1260 n!porcelain ⋅ n!metal ⋅ n!glass 3!⋅ 2!⋅ 4!

A teacher has 5 identical math books, 3 identical biology books and 4 identical physics books. In how many different ways can the teacher arrange her books on a shelf if books about the same subject must be together? The books are identical. (

1 group of biology books

+

1 group of math books

+

1 group of physics books

)! ⋅

3! 3! biology books

⋅

4! ⋅ 4! physics books

5! = 3! ⋅1 ⋅1 ⋅1 5! math books

= 6 different ways. As an extension to this example, try calculating the number of possible arrangements if books on the same subject do not have to be kept together.

EXAMPLE

46

Solution

How many three-letter words can be formed from the letters in the word NARIN if each word must contain both N’s? We have three possible sets of letters: {A, N, N}, {R, N, N} and {I, N, N}. For each set, the number of words that we can form is 3! = 3. 2! Since there are 3 sets, there are 3 ⋅ 3 = 9 possible words: ANN, NAN, NNA, RNN, NRN, NNE, INN, NIN, NNI.

EXAMPLE

47

Solution

430

I toss a coin successively 7 times. In how many ways can I get 4 heads and 3 tails? Let H represent heads and T represent tails. Then we can write the result of 7 tosses as a sequence of 7 letters. So the problem is equivalent to finding the number of seven-letter words which contain 4 H letters and 3 T letters, such as HHTHTTH, HTTHHTH or TTTHHHH. 7! 7 ⋅6 ⋅5 So the answer is = = 35 different ways. 3!⋅ 4! 3 ⋅ 2 ⋅1 Algebra 10

EXAMPLE

48

Solution

Every day, Lazar walks from his house to school. Lazar’s neighborhood has streets laid out in a grid system, as shown by the grid lines opposite. If Lazar is only allowed to walk eastward and northward along the streets, in how many different ways can he walk to school?

school

North house

East

There are 3 northward paths and 5 eastward paths. Let N stand for a northward path and E stand for an eastward North path. Then, since Lazar can walk only northward and eastward, any of Lazar’s routes can be represented by any word formed of 3 N’s and 5 E’s. For example, the word house EENEENEN represents the path opposite. 8! So the total number of ways will be = 56. 3!5!

school

East

Check Yourself 8 1. In a competition, 2 students will receive a gold medal, 3 students will receive a silver medal and 4 students will receive a bronze medal. In how many ways can the medals be awarded to 9 students? 2. A restaurant prepares a shish kebab with 5 identical pieces of meat, 3 slices of tomato and 2 identical pieces of pepper. In how many different ways can these pieces be put on the skewer 3. Esra is writing a test of multiple choice questions. Each question has 5 possible choices, and the test will include 20 questions. Esra will also prepare an answer sheet. How many different possible answer sheets can she prepare if the number of correct choices are equally distributed? Answers 1. 1260

2. 2520

3.

20! 4!4!4!4!4!

3. Circular Permutations Consider the numbers 1, 2, 3, 4 and 5 and look at two different ways of arranging them in a line:

Probability

1

2

3

4

5

Now imagine that we connect the ends of each line to make the arrangements circular. We can see that the linear arrangements are completely 4 different, but the second circular arrangement is simply a rotation of the first. Provided we do not mark a particular 3 position as the top of the circle, we can say that the two circular arrangements are identical.

5

3

4

5

1

2 2

1 2

1

3 5

4

431

Let us look at another example. In how many ways can we arrange the letters A, B and C in a line and in a circle? Linear arrangements

Circular arrangements A

ABC

ACB

BAC

BCA

CAB

CBA

B

A

C

C

B

There are 3! = 6 linear arrangements. To calculate the number of circular arrangements, we keep one letter in the same position and permute the rest. For example, let us keep A at the top and consider the linear permutations of the remaining letters B and C: A

B

C

2! = 2 ways

So there are two circular arrangements. This is a simple example of a circular permutation of three things. What about the general case for n things? Let us try to see a pattern. In the following circles, each arc represents a possible place for an object around the circle: 1st object

1 possible arc (place) (1 – 1)

2nd object

3rd object

4th object

1 possible place (2 – 1)

2 possible places (3 – 1)

3 possible places (4 – 1)

14444444444424444444444443

nth object

n – 1 possible places

Total number of permutations of three objects = 1 × 1 × 2 = (3 – 1)!

144444444444444424444444444444444443 Total number of permutations of four objects = 1 × 1 × 2 × 3 = (4 – 1)!

14444444444444444444444244444444444444444444443 Total number of permutations of n objects = 1 × 1 × 2 × 3 × ... × (n – 1) = (n – 1)!

Conclusion The total number of circular permutations of n distinct objects is (n – 1)!. 432

Algebra 10

EXAMPLE

49

Solution

EXAMPLE

50

Solution

In how many different ways can 5 girls sit around a circular table for dinner? Since there are 5 distinct people, the number of possible arrangements is (5 – 1)! = 4! = 24.

In how many different ways can a family of 2 parents and 4 children sit around a circular dining table if the parents must sit together? Four children together with the parents make 6 people. In this question we can think of the parents as a single member of the group. So there are 5 members. There are (5 – 1)! = 4! = 24 ways of sitting 5 people around a table. However, within the group of parents there will be 2! = 2 different possible linear arrangements. So the family can sit in (5 – 1)! ⋅ 2! = 24 ⋅ 2 = 48 ways around the table.

EXAMPLE

51

Solution

In how many ways can 3 Turkish, 5 Kyrgyz and 4 Turkmen diplomats be seated around a circular table if diplomats from the same country must sit next to each other? There are 3 groups of diplomats so the groups can be seated in (3 – 1)! different ways around the table. However, we must also consider the linear permutation of the diplomats in each group. So there are (3 – 1)! ⋅ 3! ⋅ 5! ⋅ 4! = 2 ⋅ 6 ⋅ 120 ⋅ 24 = 5760 different possible ways of seating the diplomats.

EXAMPLE

52

Solution

Probability

A room contains a circular table with 5 chairs and a bench for 3 people. In how many ways can 8 students be seated in the room? Let us first sit 3 students on the bench. This is a linear permutation, so there are P(8, 3) ways to do this. Then the remaining 5 students can be arranged around the circular table in (5 – 1)! different ways. So the total number of arrangements is P(8, 3) ⋅ (5 – 1)! = 8064. 433

EXAMPLE

53

Solution 1

In how many different ways can 3 girls and 3 boys be seated around a circular table with 6 chairs if no two girls must sit together? If no two girls can sit together, the students must sit alternately: B G

G

B

B G

We can seat the boys in (3 – 1)! different ways. However, if we fix the position of one boy (say B1), we can think of the rest of the seats as a linear arrangement:

B1 G

G

B

B

G

B

G

B

G

G

In this linear arrangement, the boys can be seated in 2! ways and the girls can be seated in 3! ways. So there are a total of 2! ⋅ 3! = 12 different ways to seat the students. Solution 2

Alternatively, we can make use of circular permutation. The boys can sit in (3 – 1)! different ways and the girls can also sit in (3 – 1)! different ways. However, a certain girl (say G1) can sit in three different positions with respect to a certain boy (say B1):

B1

B1

G1

B1 G1

G1

In conclusion, the children can be seated in (3 – 1)! ⋅ (3 – 1)! ⋅ 3 = 12 different ways. 434

Algebra 10

Some permutation problems ask us to arrange objects on a chain or on a circular string, for example: arrange a set of beads on a necklace, or arrange a set of keys on a key ring. In these problems we must divide the total number of circular permutations by 2 since two permutations can simply be the same arrangement viewed from the front and back.

EXAMPLE

54

Solution

In how many ways can a red bead, a blue bead, a green bead and a yellow bead be arranged on a necklace? (4 − 1)! = 3 ways. We divide by 2 because there are two different points of view, 2 from the front and from the back.

The answer is

For instance,

are the same arrangement viewed from

and

the front and the back.

Check Yourself 9 1. The construction department on Planet Zop is building a new flying saucer with 5 windows equally spaced around it. Each window will have a different color. In how many different ways can the windows be arranged? 2. In how many different ways can we put 6 keys on a key ring? 3. Three families will have a supper at a large round table. Each family has 2 parents and the families have 1, 2 and 3 children respectively. In how many different ways can all the people be seated around the table if members of the same family must sit together and the children must sit between their parents? Answers 1. (5 – 1)! = 24 Probability

2.

(6 − 1)! = 60 2

3. 192 435

EXERCISES

6 .2

A. Permutation Functions ⎛ Δ †  œ⎞ ⎟ is a permutation function 1. g = ⎜ ⎜ † œ Δ ⎟ ⎝ ⎠

8. Seven people will be in a group photograph. In how many different ways can the photograph be set up if 3 people must be in front and 4 must be at the back?

defined in Q ={€, œ, …, Δ}. Find g–1.

9. A group photograph will be taken of 5 boys and 5 ⎛1 2 3 ⎞ ⎛1 2 3 ⎞ ⎟ and g = ⎜ ⎟ are 2. The functions f = ⎜ ⎜3 1 2⎟ ⎜3 2 1 ⎟ ⎝ ⎠ ⎝ ⎠ –1 defined in A = {1, 2, 3}. Find f g

3. Two permutation functions f and g are defined in D = {1, 2, 3, 4} such that f(x) = 5 – x and g(x) = x. Write ( f g)(x) as a permutation function.

girls. Five people must be in the front and 5 people must be at the back. If the girls must sit together, in how many ways can the photograph be taken?

C. Permutations of r Elements Selected from n Elements 10. Evaluate the expressions. a. P(11, 2)

⎛a b c d e ⎞ ⎟ and g are defined in ⎜d a e c b⎟ ⎝ ⎠ ⎛a b c d e⎞ ⎟. S = {a, b, c, d, e} such that ( f g) = ⎜ ⎜b e a d c⎟ ⎝ ⎠ Find g.

4. The functions f = ⎜

c.

P( n, 4) P( n, 3)

e.

P(5, 5) P(7, 7)

b. P(8, 3) ⋅ P(5, 4) d.

P(4, 3) + P(8, 3) P(6, 3)

11. Solve the equations. B. Permutations of n Elements 5. How many different five-digit numbers can be formed by using the digits in the number 75491 once?

a. P(n + 2, 2) = 12

b. P(n, 4) = 12 ⋅ P(n, 2)

c. P(x, 2) = 72

d.

P( n +1, 2) =2 P( n, 3)

12. In how many ways can the first, second and third 6. In how many ways can a group of 7 students be seated in a row of 7 chairs if a particular student insists on being in the first chair?

places be decided in an 8-horse race if there is no tie?

13. In a computer shop, 7 out of 10 different laptop 7. In how many different ways can we name a regular pentagon using letters P, Q, R, S, T? 436

computers can be displayed in a row in the shop window. Find the number of possible window displays. Algebra 10

14. Three students will be selected from a group of 10

22. A computer is generating palindromic sequences of

students such that one student will study physics, one will study math and the other will study chemistry. If any student can study any subject, in how many ways can the students be selected?

letters using the 26 letters of the English alphabet. Each palindrome must be at least 7 letters long and at most 9 letters long with no repeated letters. How many different palindromes can be formed?

15. How many different three-digit numbers can be formed using the digits in the number 8479235 without repetition?

16. How many combinations of at most 5 letters can be formed using the letters in the word CHARITY without repetition?

17. Murat set a password on his e-mail account using the letters of the English alphabet. The password is between 1 and 3 letters long and does not contain repeated letters. How many possible passwords are there?

18. In how many different ways can 2 parents and 3 children be seated in a car if one of the parents must be in the driver’s seat and the children sit in the back while the parents sit in front?

23. A meeting room has 5 seats in the front row and 4 seats in the back row. In how many different ways can 3 friends be seated in the two rows if they must all sit together?

24. A shopkeeper has 7 different pairs of slippers. She wants to display 4 pairs of slippers on a shelf so that the pairs are kept together. Each pair can be placed facing the wall or facing out. In how many ways can she display them?

D. Permutations with Restrictions 25 . How many six-digit numbers can be formed from the digits {2, 3, 5, 6, 7, 9} if the digits 5 and 7 must be together and no digit is repeated?

19. A four-letter password will be set up of the letters in the set {A, B, C, K, V, X}. How many of the possible passwords contain repeated letters?

20. Iona, Florica, Anton and their four friends are in a group. Four students are selected from the group to line up in a row. How many of the possible rows a. do not include Iona and Florica?

26. Mulan has 4 different types of rose bulb, 6 different types of lily bulb and 7 different types of violet seed. In how many different ways can she plant them in a flower bed row if a. she can plant them in any order? b. flowers of the same type must be planted together in a row?

b. include Anton?

21. In how many different ways can 5 students be

27. Jabari, Valery and their four friends need to be

seated in a row of 8 chairs if there must be no empty chairs between them?

seated in a row. In how many ways can this be done if Jabari and Valery refuse to sit together?

Probability

437

28. In a class photograph of 9 students, 4 students

36. Find the number of eight-letter words that can be

must be in front and 5 must be at the back. In how many different ways can the students be seated if the 3 friends Mariam, Katyusha and Nataly want to sit next to each other?

formed by using each of the letters of the word AMUDERIA once if each word must begin with DER.

37. A rabbit is trained to move only down and right through the maze opposite. In how many different ways can the rabbit reach the carrot?

29. A car dealer has 3 identical red cars, 4 identical blue cars and 2 identical white cars. In how many different ways can he display them in a row?

30. How many different permutations of the letters in the word GALATASARAY are possible if the vowels must be kept together?

38.

31. How many ten-letter words can be formed from the letters in the word TAKLAMAKAN if each letter is used only once?

32. Mulan has 3 identical rose bulbs, 4 identical lily bulbs and 6 identical violet seeds. In how many different ways can she plant them in a row in a flower bed?

33. How many different seven-digit even numbers can be made by rearranging the digits in the number 2352547?

The figure above shows the road network in a part of Ankara. Nuran must go from point A to point B on foot via point C using the shortest possible route. Find the number of routes that Nuran can take.

34. How many different six-digit numbers can be

39. A child has 4 different math books, 5 identical

made by rearranging the digits in the number 335505?

dictionaries and 3 different story books. In how many different ways can she put them on a bookshelf if

35. Humuhumunukunukuapua is the native Hawaiian name for the triggerfish. Find the number of different permutations of the letters in this word. 438

a. they can be put in any order? b. books of the same kind must be kept together?

40. A restaurant offers a breakfast of 5 different items arrranged in a circle on a plate. How many different breakfast plates can be made? Algebra 10

41. Three Turkish and 5 German diplomats will have

47. 3 identical wooden horses, 4 identical cars and 2

a meeting around a circular table. In how many ways can they be seated if diplomats from the same country must sit together?

identical swans are arranged on merry-go-round. In how many ways can this be done?

48. A lion, a parrot, a wolf, a dragonfly and a rabbit 42. A hot air balloon is made up of 24 pieces of material and each piece is a different color. How many different balloons can be made?

are attending a meeting. In how many ways can they be seated around a circular table if the wolf and the rabbit must not sit together?

49. In how many different ways can Aygerim put 7 different keys on his key ring?

43. In how many ways can 5 boys and 5 girls be seated around a circular table if children of the same gender must not sit next to each other?

Mixed Problems 50. How many zeros are there at the end of › (4!)! + (5!)! + 240!?

44. A group contains 5 boys and 5 girls. In how many different ways can the boys and girls be seated around 2 different circular tables if each table has 5 seats and the boys and girls must sit at separate tables?

45. A company is making a merry-go-round with 3 different wooden horses, 2 different miniature cars and 3 different miniature planes. In how many ways can these things be arranged on the merry-go-round if the planes must be kept together?

51. In how many different ways can the letters in the word MATHEMATICS be arranged if the vowels must be kept together?

52. A ferris wheel has 22 cars and each car has 2 seats. 22 couples will ride on the wheel so that each couple rides in different car. In how many different ways can they do this?

53. Faruk has 5 different neckties. On condition that he doesn’t wear the same tie on two consecutive days, in how many ways can Faruk wear his ties an his five days at the office?

46. A Formula 1 racetrack is a simple closed curve. Seven racing cars are racing round the track. If there is no tie and without considering which car is first, second, third, etc., in how many ways can the cars be arranged on the racetrack? Probability

54. A group of 20 people includes 2 brothers. In how › many ways can this group be seated in a circle

such that there is exactly one person between the brothers? 439

When the order of the elements chosen from a set is important, we use permutation. However, order is not always important when we are choosing elements. For example, we may want to choose a certain number of people from a group to form a committee. The order of the chosen members is not important since the result is a group of people, not an ordered set. An unordered selection of elements like this is called a combination.

An r-element subset is a subset with r elements.

When we talk about a combination of n objects taken r at a time, we mean the r-element subsets of a set with n elements. We write total the number of such combinations as ⎛ n⎞ C(n, r) or ⎜ ⎟ (n, r ∈ ] and 0 ≤ r ≤ n). ⎝r ⎠ For example, if we are asked to choose two digits from the set {2, 3, 5}, we might choose {3, 5} or {5, 3}. These are the same combination. This is very different to the problem of forming a two-digit number using the digits 3 and 5 because 35 and 53 are two different outcomes.

A. COMBINATIONS OF r ELEMENTS SELECTED FROM n ELEMENTS Consider the set K = {1, 2, 3}. Let us compare the two-element combinations with the two-element permutations of the set K in a table:

K = {1, 2, 3} Combinations with 2 elements

Permutations with 2 elements

{1, 2}

12

21

{2, 3}

23

32

{1, 3}

13

31

We can see that the number of permutations with two elements is twice the number of the combinations with two elements: 2 ⋅ C(3, 2) = P(3, 2). If we now consider the three-element combinations

Combinations with 3 elements

and permutations of the set

{a, b, c}

abc

acb

bac

bca

cab

cba

A = {a, b, c, d}, we get the

{a, b, d}

abd

adb

bad

bda

dab

dba

{a, c, d}

acd

adc

cad

cda

dac

dca

{b, c, d}

bcd

bdc

cbd

cdb

dbc

dcb

following table:

440

A = {a, b, c, d} Permutations with 3 elements

Algebra 10

There are four combinations and 24 permutations. We can see that the number of permutations with three elements is 3! times the number of combinations with three elements: 3! ⋅ C(4, 3) = P(4, 3). If we repeated this exercise for two-element permutations and combinations we would find 2! ⋅ C(4, 2) = P(4, 2). We can generalize this pattern as C( n, r ) ⋅   

ways of choosing a group with r elements

r! 

= P( n, r ), which gives us the formula

ways of arranging arranging those elements

C( n, r ) =

P( n, r ) = r!

n!

( n − r)!

r!

=

n! . ( n − r )!⋅ r !

combination

Definition

Let n and r be non-negative integers such that 0 ≤ r ≤ n. element combination A subset of r elements chosen from a set of n elements is called an r-e of that set.

C(n, r) is sometimes written as Crn , ⎛ n⎞ C ⎜ ⎟ , n Cr or n Cr . ⎝r⎠ C n r is sometimes read as ‘n, choose r’.

EXAMPLE

55

Solution

EXAMPLE

56

Solution

Probability

The number of r-element combinations of a set of n elements is n! ( n, r ∈ ] and 0 ≤ r ≤ n). C(n, r) = r ! ⋅ (n – r)!

Calculate C(8, 3). By the formula, C(8, 3) =

8! 8! 8 ⋅ 7 ⋅ 6 ⋅ 5! = = = 56. 3! ⋅ (8 – 3)! 3! ⋅ 5! 3! ⋅ 5!

Evaluate C(12, 5) ⋅ C(7, 2). C(12, 5) ⋅ C(7, 2) =

12! 7! 12! 7! 12! ⋅ = ⋅ = 5! ⋅ (12 – 5)! 2!(7 – 2)! 5! ⋅ 7! 2! ⋅ 5! 5! ⋅ 2! ⋅ 5!

=

12 ⋅ 11 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 ⋅ 6 ⋅ 5! = 11 ⋅ 2 ⋅ 9 ⋅ 2 ⋅ 7 ⋅ 6 = 16632 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅ 2 ⋅ 5! 441

EXAMPLE

57

Solution

EXAMPLE

58

Solution

EXAMPLE

59

Solution

Find the number of groups of 3 students which can be chosen from a class of 10 students. The number of such groups is C(10, 3) =

10! 10 ⋅ 9 ⋅ 8 = = 120. 3!⋅ (10 − 3)! 3 ⋅ 2 ⋅1

There are 8 fruit pieces of different kinds including an apple on a tray. How many selections of 4 pieces of fruit can we make if we have to include the apple? If we have to include the apple, we need to select three pieces of fruit from the seven remaining: C(7, 3) = 35.

There are 10 players in a list. A basketball coach will choose 6 players from the list for a school team and make one of them the captain. In how many ways can the coach form the team? ⎛ ⎞ 10! = 210 ⎟. The coach can choose 6 players in 210 ways ⎜ C(10, 6) = 6!⋅ (10 − 6)! ⎝ ⎠

Additionally, any one of these six chosen players can be the captain. By the multiplication property, the coach can form the team in C(10, 6) ⋅ 6 = 1260 ways.

EXAMPLE

60

Solution

In a group of 9 children, 4 children will be given apples, another 3 children will be given oranges and the rest will be given peaches. In how many ways can these fruits be given? ⎛ 9⎞ We can choose four children from nine in ⎜ ⎟ ways and from the remaining five children we ⎝ 4⎠ ⎛ 5⎞ can choose three in ⎜ ⎟ ways. There will only be one way to choose the other two children. ⎝ 3⎠ ⎛ 9⎞ ⎛ 5⎞ So the total number of possible groupings is ⎜ ⎟ ⋅ ⎜ ⎟ ⋅ 1 = 1260. ⎝ 4⎠ ⎝ 3⎠ Note that we can also solve this problem by treating it as a permutation with some identical

elements. 442

Algebra 10

EXAMPLE

61

Solution

A cafe offers chocolate, lemon, sour cherry and vanilla flavors of ice cream. A customer can choose one, two or three scoops but the flavours must all be different. How many different possible ice creams can a customer order?

There are four types of ice cream. ⎛4⎞ ⎛4⎞ ⎛4⎞ The number of possible ice creams is ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ = 4+6+ 4 =14. ⎝1 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠ Notice that ⎛ 4⎞ ⎛4⎞ ⎛ 4⎞ ⎛ 4⎞ 4! = ⎜ ⎟. ⎜ ⎟ and ⎜ ⎟ are equal: ⎜ ⎟ = ⎝1 ⎠ ⎝ 3⎠ ⎝ 1 ⎠ 1!⋅ 3! ⎝ 3 ⎠

EXAMPLE

62

Solution

Classes 10A and 10B have 12 and 18 students respectively. A basketball team of 5 players will be formed by choosing 2 students from 10A and 3 students from 10B. How many different teams can be formed? The basketball team has five players.

⎛ 12 ⎞ We can choose 2 students from 12 students in ⎜ ⎟ ways. ⎝2⎠ ⎛ 18 ⎞ We can choose 3 students from 18 students in ⎜ ⎟ ways. ⎝3⎠ ⎛ 12 ⎞ ⎛18 ⎞ So the team can be formed in ⎜ ⎟ ⋅ ⎜ ⎟ = 66 ⋅ 816 = 53856 ways. ⎝2⎠ ⎝3⎠

EXAMPLE

63

Solution

Probability

How many three-digit numbers abc can we write which satisfy the condition c < b < a? Notice that the digits a, b and c must all be different. So any three-element set of digits {a, b, c} will be enough to form a valid number, because we can just arrange the digits to satisfy the condition. For example, the digits set is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} and from the chosen subset {3, 5, 8} we can form the number 853. So we just need to find the total number of three-digit subsets of the set of digits: C(10, 3) = 120 different numbers can be formed. 443

EXAMPLE

64

Solution

A watchmaker has 7 different jewels. He wants to choose four of them to decorate the quarters (3, 6, 9, 12) on the face of a clock. How many different decorations are possible? ⎛7⎞ The watchmaker can choose four jewels in ⎜ ⎟ different ways and set them around the quarters ⎝ 4⎠ ⎛7⎞ on the dial in 4! different ways. So the total number of possible decorations is ⎜ ⎟ ⋅ 4! = 840. ⎝ 4⎠

(Notice that we cannot use circular permutation in this problem. Can you see why?) EXAMPLE

65

Solution

A room contains a circular table with 5 chairs and a bench for 3 people. In how many ways can 8 students be seated in the room? We studied this problem in Example 64. In that solution we began by sitting three of the students on the bench. Now let us begin by sitting five students at the circular table. We can 8 choose these students in ⎛⎜ ⎞⎟ different ways. Then they can be arranged around the table in ⎝5⎠ (5 – 1)! ways. Finally, we have 3! ways for the remaining students to sit at the bench. So the 8 total number of arrangements is ⎛⎜ ⎞⎟ ⋅ 4!⋅ 3!= 56 ⋅ 24 ⋅ 6 = 8064. ⎝5⎠

EXAMPLE

66

Solution

In a queue of 5 students at the canteen, 2 students are from class A and the rest are from class B. If we know that there are 7 students in class A and 9 students in class B, find the total number of possible ways to form the queue. ⎛7⎞ In a queue, the order is significant. There are ⎜ ⎟ possible two-student groups from class A ⎝2⎠ ⎛ 9⎞ and ⎜ ⎟ possible three-student groups from class B. Those five students can form a queue ⎝ 3⎠

in 5! different ways.

⎛7⎞ ⎛9⎞ So the total number of possibilities is ⎜ ⎟ ⋅ ⎜ ⎟ ⋅ 5!= 21 ⋅ 84 ⋅120 = 211680. ⎝ 2 ⎠ ⎝ 3⎠ Remark

The number of r-element subsets of a set of n elements is equal to the number of (n – r)-element subsets: C(n, r) = C(n, n – r) (n, r ∈ ] and 0 ≤ r ≤ n). Check: C( n, n – r ) = 444

n! n! n! = = = C( n, r ). ( n – r )! ⋅ [ n – ( n – r )]! ( n – r)! ⋅ [ n – n+ r]! ( n – r)! ⋅ r! Algebra 10

EXAMPLE

67

Solution

The number of 3-element subsets of a set is equal to the number of 6-element subsets of the same set. Find the number of 7-element subsets of this set. We know that C(n, r) = C(n, n – r) and we have C(n, 3) = C(n, 6). So n = 9. ⎛9⎞ So the number of 7-element subsets of this set is ⎜ ⎟ = 36. ⎝7 ⎠

EXAMPLE

68

Solution

There are 9 students in a class. Four of them will be chosen to go on a picnic and the rest of the students will form a basketball team. In how many ways can the picnic group be chosen? Let us calculate the number of ways of establishing the basketball team, since each student is only involved in one activity: 9! 9 ⋅8 ⋅7 ⋅6 ⋅5 ⎛9⎞ = =126. ⎜5⎟ = ⎝ ⎠ 5!⋅ (9 − 5)! 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1 Note that would have found the same answer if we had begun by considering the picnic group.

EXAMPLE

69

Solution

On Planet Zop there are 7 Zozos and 5 Yoyos. The army wants to form a crew of 3 members for a spacecraft. How many different crews can be formed if there must be at least one Yoyo in the crew? ⎛ 12 ⎞ The number of groups with three members is ⎜ 3 ⎟ . This number is the sum of the ⎝ ⎠ ⎛7⎞ number of groups with only Zozos ( ⎜ ⎟ ) and the number of groups with at least one Yoyo. ⎝ 3⎠ Let us call this second group y. Then we can write ⎛ 12 ⎞ ⎛ 7 ⎞ ⎛12 ⎞ ⎛7 ⎞ ⎜ 3 ⎟ = ⎜ 3 ⎟ + y, so y = ⎜ 3 ⎟ − ⎜ 3 ⎟ which means y= 220 −35 =185. ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

So there are 185 crews which contain at least one Yoyo. As an exercise, try to develop an alternative way of solving this problem.

EXAMPLE

70

Solution 1

Probability

Esma has 4 pigeons and 5 parrots. How many different pairs containing a parrot and a pigeon can Esma choose? We can solve this question by using the multiplication principle: First bird

Second bird

⎛ 4⎞ ⎜ 1⎟ ⎝ ⎠

⎛ 5⎞ ⎜ 1⎟ ⎝ ⎠

→ 4 ⋅ 5 =20.

445

Solution 2

9 4 Alternatively, there are ⎛⎜ ⎞⎟ ways of choosing a pair of birds. Of these, ⎛⎜ ⎞⎟ pairs contain only 2 ⎝ ⎠ ⎝2⎠ ⎛5⎞ pigeons and ⎜ ⎟ pairs contain only parrots. Let d be the remaining number of mixed pairs. ⎝2⎠ Then ⎛⎜ 9 ⎞⎟ = ⎛⎜ 4 ⎞⎟ + ⎛⎜ 6 ⎞⎟ + d, so d = ⎛⎜ 9 ⎞⎟ − ⎡⎢ ⎛⎜ 4 ⎞⎟+ ⎛⎜ 5 ⎞⎟ ⎤⎥ = 36 − (6 +10) = 20. ⎝2⎠ ⎝2⎠ ⎝2⎠ ⎝ 2 ⎠ ⎣ ⎝ 2 ⎠ ⎝ 2 ⎠⎦

This second solution may seem a bit longer than the first one. However, the strategy we have just used will be helpful in other questions.

EXAMPLE

71

Solution

EXAMPLE

72

A football team is made up of 10 players plus a goalkeeper. Five more players are reserves. The team coach wants to substitute 2 team players (not including the goalkeeper) with 2 reserves and then choose 3 forward players from the resulting team. If each player can play any position, in how many ways can the coach choose the 3 forward players? ⎛ 10 ⎞ The coach can choose the players to substitute in ⎜ ⎟ different ways. He can choose the ⎝2⎠ 5⎞ ⎛ reserve players in ⎜ ⎟ different ways. Finally, he can arrange the forwards in ⎛⎜ 10 ⎞⎟ different ⎝3⎠ ⎝2⎠ 10 ⎞⎛ 5 ⎞⎛10 ⎞ ⎛ ways. So the forwards can be chosen in ⎜ ⎟⎜ ⎟⎜ ⎟ = 54000 different ways. ⎝ 2 ⎠⎝ 2 ⎠⎝ 3 ⎠

In how many different ways can 10 people be separated into 2 equivalent groups if a. one group travels to Izmir and the other group travels to Kayseri? b. the two groups play basketball together?

Solution

10 a. The group which goes to Izmir can be chosen in ⎛⎜ ⎞⎟ = 252 ways. ⎝5⎠ The rest of these ten people will go to Kayseri. So the answer is 252 ways. 10 b. There will be two teams of five players each. We can choose the first team in ⎛⎜ ⎞⎟ = 252 ⎝5⎠ ways and the rest of the people will be in the second team.

However, half of these 252 possible teams will be the same as the other half. ⎛ 10 ⎞ ⎜5⎟ ⎝ ⎠ = 126 So the ten people can be separated into two teams in ways. 2 In the same way, 15 people can be separated into three teams of five members in ⎛ 15 ⎞⎛10 ⎞⎛ 5 ⎞ ⎜ 5 ⎟⎜ 5 ⎟⎜ 5 ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠ ways. 3! In how many ways could 12 students be separated into four equal teams? 446

Algebra 10

EXAMPLE

73

Solution

Derive a formula that gives the number of diagonals in a convex polygon with n sides. A polygon with n sides has n vertices A diagonal is a line segment which joins two non-adjacent vertices. The total number of line segments which join any of the vertices is n! n ⋅ ( n − 1) ⎛ n⎞ determined by the two-element subsets of all the vertices: ⎜ ⎟ = = . 2 ⎝ 2 ⎠ 2! ⋅ ( n − 2)!

However, line segments which join two adjacent vertices are not counted as diagonals. They are the sides. If we subtract n sides from the first formula, the formula for the number of diagonals is n ⋅ ( n − 1) n ⋅ ( n − 3) ⎛ n⎞ −n = . ⎜2 ⎟ − n = 2 2 ⎝ ⎠

For example, in a hexagon there are

EXAMPLE

74

Solution

6 ⋅ (6 − 3) = 9 diagonals. 2

I toss a coin successively 7 times. In how many ways can I get 4 heads and 3 tails? We have already solved this question as a permutation in Example 59. Now we can solve it as a combination. Let the numbers 1, 2, 3, 4, 5, 6, 7 represent each toss of the coin. Then any four-element subset chosen from this set will represent a group of outcomes in which the coin is heads. For example, 1, 2, 4, 6 means heads on the first, second, fourth and sixth toss. 2, 3, 4, 5 is another possibility. 7 There are ⎛⎜ ⎞⎟ = 35 such groups, so can get the result in 35 ways. ⎝4⎠

EXAMPLE

75

Solution

A canteen has 2 circular tables with 4 seats each and a bench with 5 seats. In how many different ways can 13 students chosen from a class of 15 students sit in the canteen? ⎛ 15 ⎞ We can choose 13 students in ⎜ 13 ⎟ different ways. ⎝ ⎠ ⎛ 13 ⎞ From these students, ⎜ ⎟ students will sit at the first circular table and ⎝4⎠ sit at the second circular table. These students can be seated in 3! and 3!

⎛9⎞ ⎜ 4 ⎟ students will ⎝ ⎠ ways respectively.

The remaining five students can sit on the bench in 5! different ways. So the total number of possible arrangements is a product: ⎛ 15 ⎞ ⎛13 ⎞ ⎛ 9 ⎞ ⎜ 13 ⎟ ⋅ ⎜ 4 ⎟ ⋅ ⎜ 4 ⎟ ⋅ 3! ⋅ 3! ⋅ 5!=105 ⋅715 ⋅126 ⋅6 ⋅6 ⋅120 = 40 864 824 000. ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ Probability

447

EXAMPLE

76

There are 13 bulbs in a box. Five of the bulbs are defective. We will select a set of 4 bulbs from the box. In how many ways can we do this if a. none of the bulbs must be defective? b. we want at least half of the set to be defective?

Solution

8 a. There are eight working bulbs, so we must select four bulbs from eight bulbs: ⎛⎜ ⎞⎟ = 70 ⎝4⎠ possible sets.

b. There are four bulbs in the set. If the set contains at least two defective bulbs, there must be 2 or 3 or 4 defective bulbs and 2 or 1 or zero working bulbs in each set: ⎛ 5 ⎞⎛ 8 ⎞ ⎛ 5 ⎞⎛ 8 ⎞ ⎛ 5 ⎞⎛ 8 ⎞ ⎜ 2 ⎟⎜ 2 ⎟ + ⎜ 3 ⎟⎜ 1 ⎟ + ⎜ 4 ⎟⎜ 0 ⎟ =(10 ⋅ 28)+(10 ⋅8)+(5 ⋅1) = 365. ⎝ ⎠⎝ ⎠ ⎝  ⎠⎝ ⎠ ⎝  ⎠⎝ ⎠     groups with 2 defective bulbs

EXAMPLE

77

Solution

EXAMPLE

78

groups with 3 defective bulbs

groups with 4 defective bulbs

There are 11 points on a circle. How many triangles can we form using any three of these points as the vertices? Any three points on the circle are not collinear. So any three points will 11 make a triangle. So the total number of triangles is ⎛⎜ ⎞⎟ =165. ⎝3⎠ Seven points are given as shown in the adjacent figure. a. How many lines can be draw which pass through at least two of the points? b. How many triangles can be formed using the points as vertices?

Solution

a. Two lines are already given. There are three collinear points on the top line and four collinear points on the bottom line. Other lines can pass through one of the top and one of the bottom points. There are 3 ⋅ 4 = 12 such lines. Including the top and bottom line, there are 12 + 2 = 14 possible lines. b. For any triangle we want to draw, there are two cases: Case 1: A side is on the upper row and the vertex is a point on the lower row. Since two points determine a 3 side, there can be ⎛⎜ ⎞⎟ ⋅ 4 =12 such triangles. 2 ⎝ ⎠ Some of them are shown in the figure.

448

Algebra 10

Case 2: A side is on the lower row and the vertex is on 4 the upper row. There are ⎛⎜ ⎞⎟ ⋅ 3 = 18 such triangles. ⎝2⎠ Some of them are shown in the figure. In conclusion, we can form 12 + 18 = 30 triangles.

EXAMPLE

79

There are 4 permanent members and 9 elected members on a company’s board of directors. For a decision to be passed, there must be at least 8 votes in favor of the decision and all of the permanent members must vote in favor. In how many ways can a decision be passed by the board, assuming that all members vote?

Solution

For a decision to be passed, all the permanent members must vote in favor. This is possible ⎛4⎞ in ⎜ ⎟ = 1 way. ⎝4⎠ But this is not enough. At least four more votes are necessary since at least eight votes are needed in favor. These four or more votes can be provided by the nine elected members of the board in ⎛⎜ 9 ⎞⎟ , ⎛⎜ 9 ⎞⎟ , ⎛⎜ 9 ⎞⎟ , .... or ⎛⎜ 9 ⎞⎟ different ways. ⎝4⎠ ⎝5⎠ ⎝6⎠ ⎝9⎠ 9 9 9 9 9 9 So there are ⎛⎜ ⎞⎟ + ⎛⎜ ⎞⎟ + ⎛⎜ ⎞⎟ + ⎛⎜ ⎞⎟+ ⎛⎜ ⎞⎟+ ⎛⎜ ⎞⎟ ways in which at least four of the nine ⎝ 4 ⎠ ⎝ 5 ⎠ ⎝ 6 ⎠ ⎝7 ⎠ ⎝ 8 ⎠ ⎝9 ⎠ elected members can vote in favor of the decision. As a result, a decision can be passed in

⎛4⎞ ⎛9⎞ ⎛4⎞ ⎜ ⎟ ⋅ ⎜ ⎟+⎜ ⎟ ⋅ ⎝4⎠ ⎝4⎠ ⎝4⎠

⎛9 ⎞ ⎛4 ⎞ ⎜ ⎟+ ⎜ ⎟ ⋅ ⎝5 ⎠ ⎝4 ⎠

⎛ 9 ⎞ ⎛ 4 ⎞ ⎛9 ⎞ ⎛ 4 ⎞ ⎜ ⎟ + ⎜ ⎟ ⋅ ⎜ ⎟+ ⎜ ⎟ ⋅ ⎝ 6 ⎠ ⎝ 4 ⎠ ⎝7 ⎠ ⎝ 4 ⎠

⎛9 ⎞ ⎛ 4 ⎞ ⎜ ⎟+ ⎜ ⎟ ⋅ ⎝8 ⎠ ⎝ 4 ⎠

⎛9 ⎞ ⎜ ⎟ ⎝9 ⎠

⎛ 4 ⎞ ⎡⎛ 9 ⎞ ⎛ 9 ⎞ ⎛ 9 ⎞ ⎛ 9 ⎞ ⎛ 9 ⎞ ⎛ 9 ⎞ ⎤ = ⎜ ⎟ ⋅ ⎢⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟+ ⎜ ⎟+ ⎜ ⎟ ⎥ ⎝ 4 ⎠ ⎣⎝ 4 ⎠ ⎝ 5 ⎠ ⎝ 6 ⎠ ⎝ 7 ⎠ ⎝ 8 ⎠ ⎝ 9 ⎠ ⎦ =1 ⋅ (126+126+84+ 36+9+1) = 382 ways.

EXAMPLE

80

Solution Probability

⎛ n ⎞ ⎛ n − 1⎞ ⎛ n − 1 ⎞ Prove that ⎜ ⎟ = ⎜ ⎟+ ⎜ ⎟ , where n, r ∈ ` with r ≤ n. ⎝ r ⎠ ⎝r −1⎠ ⎝ r ⎠

This solution is left as an exercise for you. 449

number of subsets of a set

Theorem

The total number of subsets of a set with n elements is 2n: ⎛ n⎞ ⎛ n⎞ ⎛ n⎞ ⎛n⎞ n ⎜ 0 ⎟ + ⎜ 1 ⎟ + ⎜ 2 ⎟ + ⋅ ⋅ ⋅ + ⎜ n ⎟ = 2 , n ∈ ] (0 ≤ n). ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Proof

For a given subset there are two possibilities for each element of the main set: either it is in the subset or not in the subset. If the n elements are a1, a2, a3, ..., an we can construct the following table: a1

a2

a3

...

an

2

2

2

...

2

.

 

n elements

So the total number of subsets is 2n. EXAMPLE

81

Solution EXAMPLE

82

Solution

Find the total number of subsets of the set K = {a, b, c, d, e, f, g, h}. Since the set K has 8 elements, the total number of subsets is 28 = 256. How many subsets of the set P = {1, 2, 3, 4, 5, 6, 7} contain 2 or 6 or both 2 and 6? The total number of subsets of P is 27. The subsets which do not contain either of the digits 2 and 6 are in fact the subsets of the set {1, 3, 4, 5, 7}, which has 25 subsets. So the number of subsets that contain 2 or 6 or both is 27 – 25 = 128 – 32 = 96.

Check Yourself 10 1. P(n, 3) = (n + 1) ⋅ C(n, 3) is given. Find n. 2. Find the number of subsets of the set {a, b, c, d, e} which contain at least 3 elements. 3. There are 12 people in a room. Each person shakes hands with all the other people. How many handshakes are there? 4. In how many different ways can Hunfrid distribute 9 different toys among 3 children so that each child gets 3 toys? 5. A box holds 7 red cards and 5 green cards. How many different groups of 6 cards can be selected from the box if the selection must contain at least 3 red cards? Answers 1. 5

450

⎛5⎞ ⎛5⎞ ⎛5⎞ 2. ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ ⎜ 3⎟ ⎜ 4⎟ ⎜ 5⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

3. 66

⎛ 9 ⎞⎛ 6 ⎞⎛ 3 ⎞ 4. ⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ 3 ⎟⎜ 3 ⎟⎜ 3 ⎟ ⎝ ⎠⎝ ⎠⎝ ⎠

5. 812 Algebra 10

EXERCISES

6 .3

A. Combinations of r Elements Selected from n Elements 1. Evaluate the expressions.

combination for an operation in a program. For this purpose, he will use two of the keys Shift, Ctrl or Alt together with one of 26 letters. How many different key combinations can he choose from?

b. C(6, 2) + C(8, 3) P(7,4) C(7,4)

d. ⎛⎜ 9 ⎞⎟ + ⎛⎜ 9 ⎞⎟ + ⎛⎜ 9 ⎞⎟ + ⎛⎜ 9 ⎞⎟ +...+ ⎛⎜ 9 ⎞⎟ ⎝ 0 ⎠ ⎝1 ⎠ ⎝ 2 ⎠ ⎝ 3 ⎠ ⎝9⎠ e. ⎛⎜ 13 ⎞⎟ + ⎛⎜13 ⎞⎟ + ⎛⎜13 ⎞⎟ +...+ ⎛⎜13 ⎞⎟ ⎝0⎠ ⎝2⎠ ⎝4⎠ ⎝12 ⎠

c. C(n, 2) + C(n, n – 2)

b. ⎛⎜ n ⎞⎟ + 20 = ⎛⎜ 8 ⎞⎟ ⎝2 ⎠ ⎝3⎠

c. ⎛⎜ n +1⎞⎟ = ⎛⎜ n − 1 ⎞⎟ +17 ⎝ n −1⎠ ⎝ 2 ⎠

Assuming that an arc is represented by two letters and the order of the letters is important, how many different arcs can we define?

G = {k, l, m, n, r}.

5. How many subsets of at least 4 elements does the set K = {, z, œ, …, €, X,

the set {2, 3, 4, 5, 6,}. How many three-digit numbers abc can be formed in this way such that a < b < c?

13. In an aquarium there are 5 different white fish, 7

4. List all the three-element subsets of the set

} have?

6. How many of the three-element subsets of the set H = {a, b, c, d, e, f} include the letter e?

7. In how many different ways can one novel, one biography and one poetry book be chosen from 3 novels, 4 biographies and 5 poetry books? Probability

with everybody else. How many handshakes are there?

12. Three different digits a, b and c are chosen from

3. Solve the equations. a. C(n, 2) = 15

10. In a group of 10 people, everybody shakes hands

11. A, B, C, D and E are 5 distinct points on a circle.

2. Simplify the expressions. C( n, 3) a. 2C(4, 2) b. P( n, 2)

clean her house. How many different groups can she choose?

9. A computer programmer wants to set a key

a. C(4, 2)

c.

8. Snow White wants to choose 3 of the 7 dwarfs to

different goldfish and 4 different black fish. Four of the fish must be moved to a different aquarium. In how many ways can this be done?

14. Joo-Chan is making a 6 × 8 crossword puzzle in which he wants to put 12 black squares. How many different crossword patterns can he make?

15. In an attempt to travel around the world, you must visit 36 out of 40 predetermined cities which are all located near the Equator. In how many ways can you choose the group of cities you will not visit? 451

16. There are 11 children of different heights in a

23. In a group of 7 people, 3 people have a driver’s

class. We want to choose any 4 children and line them up according to their height. In how many different ways can this be done?

license. How many different groups of 5 people can be chosen to travel in a car if the group must have at least one driver?

17. Junko must choose two courses from the list M,

24. A research group of 4 people will be chosen from

N, Q, P and R for this semester. How many choices does he have if the lessons for M and Q are given at the same time?

a board of 3 professors and 5 research managers. If the group must include at least one professor, how many different groups can be chosen?

18. A ship has 12 different barrels of oil. Eight of

25. An urn contains 5 green marbles and 7 red marbles.

them will be unloaded at a port. Of these 8 barrels, 3 barrels will be painted. In how many ways can the 3 barrels be chosen?

19. In a class of 55 students, the number of possible pairs which can be formed by the girls is equal to the total number of boys in the class. Find the number of boys in this class.

20. Two basketball teams of 5 players each are playing a match. We want to choose 2 players from each team and seat them around a circular table after the match. In how many different ways can this be done?

21. In classes 9A, 9B and 9C there are 12 students,

A student takes a sample of 7 marbles from the urn. How many different samples are possible if a. the sample must contain exactly three green marbles? b. the sample must include at least five red marbles?

26. There are 5 candidates for the presidency of a school committee. A voter is allowed to vote for at most 3 candidates. In how many different ways can a student prepare her vote?

27. There are 7 points on a circle. How many pentagons can we form by joining 5 of the points?

28 . The figure below shows nine points on the › perimeter of a triangle.

10 students and 9 students respectively. A football team will be formed by choosing 5 students from 9A, 4 students from 9B and 2 students from 9C. How many different football teams can be formed?

22. Ali and his 7 friends are in group. Three people are chosen from the group to form a committee. How many of the possible committees include Ali? 452

a. How many different lines can be drawn which pass through at least two of the given points? b. Using the given points as vertices, how many different triangles can be constructed? Algebra 10

29. How many parallelograms are formed by the intersection of the parallel lines in the figure below? d1

35. Find the number of different non-negative

››

integer solutions to a + b + c + d ≤ 2006.

d 2 d 3 d4 l1 l2

l3 l4

l5 l6 l7

30. a, b, c and d are 4 digits such that a < b < c < d.

36. Fifty identical marbles are to be distributed among 6 children. If each child must get at least one marble. How many possible distributions are possible?

37. Twenty-five identical red marbles and 33 identical yellow marbles are to be distributed among 5 children so that each child should take any kind, how many different distributions are there?

If we write all the possible four-digit numbers abcd in ascending order, in which position will the number 3458 appear?

38. A chocolate shop sells 12 different types of

31. A student must select 4 out of 6 essay questions

chocolate. A customer wants to buy a selection of 20 chocolates. How many different selections are possible?

and 10 out of 15 multiple-choice questions in his mid-term examination. In how many different ways can the student select the 14 questions?

32. A company has 4 workers who can build a wall, 5 workers who can paint a wall and 2 workers who can do both. A job requires 3 builders and 3 painters. In how many ways can the company choose the workers for the job?

B. Combinations with Identical Elements 33. Find the number of different positive integer solutions to a + b + c + d + e = 27.

34. Find the number of different non-negative integer solutions to a + b + c + d = 18. Probability

39. A teacher wants to distribute 20 identical gifts among his 16 students. In how many different ways can he do this if a. he must give each student at least one gift? b. there is no restriction on the number of gifts a student can receive?

Mixed Problems 40. If P(n + 1, 4) = 40 ⋅ C(n –1, 2), find n.

41. In how many ways can 4 people be seated on a sofa for 6 people?

42. In how many ways can 4 out of 6 people be seated on a sofa for 4 people? 453

43. In how many ways can 3 out of 5 people be seated on 5 chairs in a row?

44. In how many ways can 5 out of 7 people be seated around a circular table?

45 . In how many ways can we arrange 4 out of 5 different math books and 4 out of 6 physics books on a shelf if a. the books can be put in any order? b. books on the same subject must be together?

48. Five countries each send 4 diplomats to an international meeting. After the meeting, 2 diplomats from each country are chosen for another meeting around a circular table. How many different seating arrangements are possible around the table if diplomats from the same country must sit together?

49. Jandos has 6 rubber stamps and a circular holder that can hold 12 rubber stamps. In how many ways can Jandos put his stamps in the holder?

46. A software package consists of 7 programs

50. A scuba diving school has 7 teachers and 2

including an antivirus program and a word processor. Ali’s computer has 3 available hard drives. Ali must install 4 different programs from the package, two of which must be the antivirus program and the word processor. In how many different ways can Ali do this?

student teachers. Five of these people will dive in a coral reef and the rest will teach in a classroom. In how many ways can the groups be arranged if the two student teachers must be kept together?

51. Ali, Veli and their 6 friends are in a sports club. A 47. A restaurant offers a self-service buffet lunch. For lunch there are 3 choices of soup, 6 choices of main meal and 5 choices of dessert. Customers can also make a salad from 3 out of 18 things served at the salad bar. In how many different ways can a customer make a four-course meal of soup, a main meal, salad and dessert? 454

teacher wants to choose a group of 3 people from the club so that Ali is chosen but Veli is not chosen. In how many ways can she do this?

52. Four women and 7 men are at a restaurant. One of the men is married to one of the women. In how many different ways can 5 people from this group be seated around a circular table if the group must contain the couple and the couple must sit together? Algebra 10

A. PASCAL’S TRIANGLE AND BINOMIAL EXPANSION Look at the picture opposite. A mouse is moving from

1

circle to circle from left to right across the page. After 1

each circle there are two ways for the mouse to proceed: right and up or right and down. The

1

number in each circle is the number of ways in which the mouse can

4

1

reach that circle.

Start

The numbers in the circles

3

1

2

6

1

show a pattern which is known

3

as Pascal’s triangle. This triangle

1

has many interesting properties. To

4

understand them, let us move the

1

triangle to an upright position. 1

Row

Sum

0 1

1

2

1

3

1

4

1

5

1

6

1

7

1

8

1 ↓

Probability

1

1

8

3

5 6 28

6

15 56

4 5

35 70

32

1 6

21 56

16

1

15

20

8

1

10

35

4

1 3

10

21

7

2

4

2

1

7 28

64

1

128

1 8

256

1 ↓

455

First notice that each row begins and ends with 1. Secondly, notice that the sum of any two consecutive terms in a row gives us the term between them on the next row. For instance, the number 15 marked in red in the sixth row is the sum of the 10 and 5 located above it. We can extend the triangle infinitely downwards by using this rule. Notice also that the first row is row zero. For convenience, when we count the positions of the numbers in each row we also begin with zero (not 1). For example, number 21 marked in green is the second entry (not the third entry) in the seventh row. The entries in Pascal’s triangle are related to the coefficients of the expansion of a binomial with a non-negative integer power. To understand the relationship, look at some binomial expansions with the first few powers and notice their coefficients. ( a + b )0 = 1 ( a + b)1 = a + b = 1a + 1b ( a + b)2 = a2 + 2 ab + b2 = 1a2 + 2ab + 1b2 ( a − b )3 = a3 − 3a2 b + 3ab 2 − b 3 = 1a3 − 3a2 b + 3ab2 − 1b3 ( x + 2 y)4 = x4 +8 x3 y + 24 x2 y2 + 32 xy3 +16 y4 = 1x4 + 4x3(2 y)+ 6 x2(2 y)2 + 4x(2 y)3 + 1(2 y)4

We can see that the coefficients in each expansion are the same as the entries in the corresponding row of Pascal’s triangle. Remark

The expansion of a binomial expression to the nth power has the following properties: ² There are n + 1 terms in the expansion. ² The coefficients of the terms in the expansion correspond to the entries in the nth row of Pascal’s triangle. ² The power of the first term in the binomial expression begins at n in the expansion and decreases by 1 in each term down to zero. ² The power of the second term in the binomial expression begins at zero in the expansion and increases by 1 in each term up to n. ² In the expansion of (x + y)n, the sum of the exponents of x and y in each term is n. A constant term in an expression is a term that does not change with the variable.

456

² The sum of the coefficients of an expansion can be found by substituting 1 for each variable in the binomial expression. ² If the binomial expression is a polynomial then substituting zero for each variable in the binomial expression gives us the constant term of the expansion. Algebra 10

EXAMPLE

83

Solution EXAMPLE

84

Solution

How many terms are there in the expansion of (x + y)12? Since the power is 12 (n = 12) there will be n + 1 = 13 terms.

Expand (2x + y)6. The primary coefficients of the terms in the expansion will be the entries in the sixth row of Pascal’s triangle: 1, 6, 15, 20, 15, 6, 1. The first term of the binomial is 2x. To avoid any mistakes, let us keep 2x in parantheses to begin with: (2x + y)6 = (2x)6 + 6(2x)5y + 15(2x)4y2 + 20(2x)3y3 + 15(2x)2y4 + 6⋅2xy5 + y6 = 64x6 + 6(32x5)y + 15(16x4)y2 + 20(8x3)y3 + 15⋅4x2y4 + 6⋅2xy5 + y6 = 64x6 + 192x5y + 240x4y2 + 160x3y3 + 60x2y4 + 12xy5 + y6.

EXAMPLE

85

Solution

Expand (x2 – 3y)4. We will use the entries in the fourth row as the coefficients: 1, 4, 6, 4, 1. If we write consider (x2 – 3y)4 as (x2 + (–3y))4 then (x2 – 3y)4 = (x2)4 – 4(x2)3(3y) + 6(x2)2(3y)2 – 4(x2)(3y)3 + (3y)4 = x8 – 12x6y + 6x49y2 – 4x227y3 + 81y4 = x8 – 12x6y + 54x4y2– 108x2y3 + 81y4.

EXAMPLE

86

Solution

7

Expand ⎛⎜ n + 1 ⎞⎟ . n⎠ ⎝ Use the seventh row: 7

2

1⎞ ⎛ 7 6 1 5 ⎛1 ⎞ 4 ⎜ n + ⎟ = n +7 n + 21n ⎜ ⎟ + 35 n n⎠ n ⎝ ⎝n ⎠ = n7 +7 n6

87

Solution Probability

4

⎛1 ⎞ 2 ⎜ ⎟ + 21n ⎝n ⎠

5

6

7

⎛1 ⎞ ⎛1 ⎞ ⎛1 ⎞ ⎜ ⎟ +7 n ⎜ ⎟ + ⎜ ⎟ ⎝n ⎠ ⎝n ⎠ ⎝n ⎠

1 1 1 1 1 1 1 + 21n5 2 + 35 n4 3 + 35 n3 4 + 21n2 5 +7 n 6 + 7 n n n n n n n

= n7 +7 n5 + 21n3 + 35 n +

EXAMPLE

3

⎛1 ⎞ 3 ⎜ ⎟ + 35 n ⎝n ⎠

35 21 7 1 + + + . n n3 n5 n7

Find the sum of the coefficients in the expansion of (3x + y2)6. If we substitute 1 for each variable in (3x + y2)6 we get (3 ⋅ 1 + 12)6 = 46 = 4096. 457

EXAMPLE

88

Solution

Find the constant term in the expansion of (7x + 3)5. Substitute zero for each variable in (7x + 3)5: (7 ⋅ 0 + 3)5 = 35 = 243.

Check Yourself 11 Expand the binomials. 4

1. (3x + y)

⎛ 1⎞ 2. ⎜ x – ⎟ ⎝ x⎠

3

3. (2 – ñ2)5

Answers 1. 81x4 + 108x3y + 54x2y2+12xy3 + y4

B.

3 2. x – 3x +

3 1 – x x3

3. 232 – 164ñ2

FINDING BINOMIAL TERMS USING COMBINATION A laboratory mouse has been exposed to five types of virus. A scientist wishes to find out how many viruses are now present in the mouse. In how many ways could the mouse have been infected? Infected with no viruses: C(5,0) = 1 (mouse is clean) Infected with 1 type of virus:

C(5,1) = 5

Infected with 2 types of virus:

C(5,2) = 10

Infected with 3 types of virus:

C(5,3) = 10

Infected with 4 types of virus:

C(5,4) = 5

Infected with 5 types of virus:

C(5,5) = 1.

Can you see the similarity between the number of combinations and the entries in the fifth row of Pascal’s triangle? Perhaps one of the most interesting characteristics of Pascal’s triangle is its relationship with combination. We can describe this relationship simply: entry number r in row n is the number of subsets of r elements which can be taken from a set with n elements. This gives us another interesting characteristic of Pascal’s triangle: the sum of the terms in the nth row of the triangle is 2n (can you see why?). The symmetrical property of Pascal’s triangle can also be related to the combination rule ⎛ n⎞ ⎛ n ⎞ ⎜ ⎟=⎜ ⎟ ( n, r ∈ ] and 0 ≤ r ≤ n). ⎝ r ⎠ ⎝n − r⎠ 458

Algebra 10

For example, the third entry in the eighth row of the triangle is the same as the fifth entry in the same row since C(8, 3) = C(8, 5) = 56. Let us now redraw Pascal’s triangle using combination: Row ⎛0⎞ ⎜0⎟ ⎝ ⎠

0

⎛2⎞ ⎜1 ⎟ ⎝ ⎠

⎛2⎞ ⎜0⎟ ⎝ ⎠

2

⎛ 3⎞ ⎜1 ⎟ ⎝ ⎠

⎛ 3⎞ ⎜0⎟ ⎝ ⎠

3 ⎛ 4⎞ ⎜0⎟ ⎝ ⎠

4

⎛6⎞ ⎜0⎟ ⎝ ⎠

⎛6⎞ ⎜ 1⎟ ⎝ ⎠

⎛3⎞ ⎜2⎟ ⎝ ⎠

⎛5⎞ ⎜2⎟ ⎝ ⎠ ⎛6⎞ ⎜2⎟ ⎝ ⎠

⎛2⎞ ⎜2⎟ ⎝ ⎠

⎛4⎞ ⎜2⎟ ⎝ ⎠

⎛ 4⎞ ⎜ 1⎟ ⎝ ⎠ ⎛ 5⎞ ⎜ 1⎟ ⎝ ⎠

⎛5⎞ ⎜0⎟ ⎝ ⎠

5 6

⎛ 1⎞ ⎜ 1⎟ ⎝ ⎠

⎛1 ⎞ ⎜0⎟ ⎝ ⎠

1

⎛ 3⎞ ⎜ 3⎟ ⎝ ⎠ ⎛ 4⎞ ⎜ 3⎟ ⎝ ⎠

⎛ 5⎞ ⎜ 3⎟ ⎝ ⎠ ⎛6⎞ ⎜ 3⎟ ⎝ ⎠

⎛ 4⎞ ⎜ 4⎟ ⎝ ⎠ ⎛5⎞ ⎜ 4⎟ ⎝ ⎠

⎛6⎞ ⎜ 4⎟ ⎝ ⎠

⎛ 5⎞ ⎜ 5⎟ ⎝ ⎠ ⎛6⎞ ⎜5⎟ ⎝ ⎠

⎛6⎞ ⎜6⎟ ⎝ ⎠

Using the above triangle, we can generalize the expansion of a binomial to any power n as ⎛ n⎞ ⎛ n⎞ ⎛n⎞ ⎛ n ⎞ n −1 ⎛ n ⎞ n ( x + y)n = ⎜ ⎟ xn + ⎜ ⎟ xn −1y + ⎜ ⎟ x n − 2 y2 +. . .+ ⎜ ⎟ xy + ⎜ n ⎟y 0 1 2 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ n −1⎠ ⎝ ⎠

.

Notice that the coefficient 1 in the first and last terms of the expansion is obtained from and ⎛⎜ n ⎞⎟ respectively. ⎝ n⎠

EXAMPLE

89

Solution

⎛ n⎞ ⎜0 ⎟ ⎝ ⎠

Expand (x + y)6 using combination. ⎛6⎞ ⎛6 ⎞ ⎛6 ⎞ ⎛6 ⎞ ⎛6 ⎞ ( x + y)6 = x6 + ⎜ ⎟ x6 −1y + ⎜ ⎟ x6 −2 y2 + ⎜ ⎟ x6 −3 y3 + ⎜ ⎟ x6 −4 y4 + ⎜ ⎟x6 −5 y5 + y6 1 2 3 4 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝5 ⎠

= x6 + 6x5y + 15x4y2 + 20x3y3 + 15x2y4 + 6xy5 + y6 Probability

459

EXAMPLE

90

Solution

Expand (2a – b)5. ⎛5⎞ ⎛5 ⎞ ⎛5 ⎞ (2a − b )5 = (2 a)5 + ⎜ ⎟(2 a)5 −1( − b)+ ⎜ ⎟ (2 a)5 −2( −b) 2 + ⎜ ⎟(2 a) 5 −3( −b) 3 + 1 2 ⎝ ⎠ ⎝ ⎠ ⎝3 ⎠

⎛5 ⎞ 5 −4 4 5 ⎜4 ⎟(2 a) ( −b) +( −b) ⎝ ⎠

= 32 a5 − 5 ⋅ 16 a4 b +10 ⋅ 8 a3 b2 −10 ⋅4 a2 b3 +5 ⋅2 ab4 − b5 = 32 a5 − 80 a4 b +80 a3 b2 − 40 a2 b3 +10 ab 4 − b5

EXAMPLE

91

Solution

7

Expand ⎛⎜ x + 1 ⎞⎟ . x⎠ ⎝ 7 ⎛7 ⎞ ⎛7 ⎞ 1⎞ ⎛ 7 6 ⎛1 ⎞ 5 + x ⎜ ⎟ =( x ) + ⎜⎜ ⎟⎟ ( x ) ⎜ ⎟+ ⎜⎜ ⎟⎟( x ) x⎠ x 1 2 ⎝ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

2 ⎛ 1 ⎞ ⎛7 ⎞ 4 ⎜ ⎟ + ⎜⎜ ⎟⎟( x ) ⎝ x ⎠ ⎝3 ⎠

3 ⎛ 1 ⎞ ⎛7 ⎞ 3 ⎜ ⎟ + ⎜⎜ ⎟⎟( x ) ⎝ x ⎠ ⎝4 ⎠

4

⎛1 ⎞ ⎜ ⎟ ⎝x ⎠

5 6 7 ⎛7⎞ ⎛7 ⎞ ⎛ 1⎞ ⎛1⎞ ⎛1⎞ + ⎜ ⎟ ( x )2 ⎜ ⎟ + ⎜ ⎟( x ) ⎜ ⎟ + ⎜ ⎟ ⎜5⎟ ⎜6 ⎟ ⎝ x⎠ ⎝x⎠ ⎝x⎠ ⎝ ⎠ ⎝ ⎠

= x7 +7 x6

1 1 1 1 1 1 1 + 21 x5 2 + 35 x4 3 + 35 x3 4 + 21 x2 5 +7 x 6 + 7 x x x x x x x

= x3 x +7x2 + 21 x +

35 35 x 21 7 x 1 + + 4+ 6 + 7 x x3 x x x

The relation between combination and Pascal’s triangle helps us to calculate any particular term in a binomial expansion without writing out the entire expansion. For example, suppose that we are asked to find the third term in the expansion of (x – 2y)3. Using our knowledge of the properties of binomial expansion, we can say that 2y will have exponent 2 in this term and x will have exponent 1 since the sum of the exponents must be 3 3. Now we only need to find the coefficient, which we can calculate as ⎛⎜ ⎞⎟ = 3. ⎝2⎠ ⎛3⎞ 2 2 2 So the third term is ⎜ ⎟ x(2 y) = 3x ⋅ 4y = 12 xy . We can easily check this against the full ⎝2⎠ expansion: (x – 2y)3 = x3 – 6x2y + 12xy2 – 8y3. We can formulize our findings as follows: 460

Algebra 10

Remark

n ⎞ n−( r −1) r −1 ⎛ n ⎞ n− r+1 r−1 The rth entry in the expansion of (x + y)n is ⎛⎜ y =⎜ y . ⎟x ⎟x ⎝ r − 1⎠ ⎝ r − 1⎠

EXAMPLE

92

Solution

EXAMPLE

93

Solution

Find the ninth term in the expansion of (a + b)12. 12 ⎞ 12 − 9 +1 9 −1 Substitute n = 12 and r = 9 in the formula: ⎛⎜ y = 495a4 b8 . ⎟x − 9 1 ⎝ ⎠

Find the sixth term in the expansion of (2x + y)9. We will use n = 9 and r = 6. So r – 1 = 5 and the sixth term is ⎛9⎞ 9 −5 5 4 5 4 5 ⎜ 5 ⎟ (2 x) y = 126 ⋅16 x y = 2016 x y . ⎝ ⎠

EXAMPLE

94

Solution

Probability

What is the coefficient of the fourth term in the expansion of (2x – 4y)7? 7 r = 4 means r – 1 = 3. So the fourth term is ⎛⎜ ⎞⎟ (2 x)7−3( −4 y)3 = 35 ⋅16 x4( −64) y3. From this ⎝ 3⎠ we can calculate the coefficient to be –35840. 461

EXAMPLE

95

Solution

EXAMPLE

96

Solution

Find the middle term in the expansion of (2x3 + 3y)8. Since there are nine terms in the expansion of a binomial to the eighth power, the middle 12 4 ⎛8⎞ 3 8 −4 4 term will be the fifth term. So the middle term is ⎜ ⎟ (2 x ) ( −3y) = 90720 x y . ⎝4⎠

27xy3 is a term in the expansion of (ax + 2y)4. Find a. The exponent of y is 3 and this is one less than the order of the term. Using r – 1 = 3 in the ⎛4⎞ 4−3 3 formula gives the term as ⎜ ⎟ ( ax) (2 y) . ⎝3⎠ If we equate this with the given term we have

⎛4⎞ 4−3 3 7 3 ⎜ 3 ⎟ ( ax) (2 y) = 2 xy ⎝ ⎠

4(ax)4–3(2y)3 = 27xy3 4(ax)23y3=27xy3 25(ax)y3 = 27xy3. So a is 22 = 4.

EXAMPLE

97

Solution

10

3 What is the constant term in the expansion of ⎛⎜ 2 + x ⎞⎟ ? 2 2 ⎠ ⎝x

⎛ 10 ⎞ ⎛ 2 ⎞10 − r +1 ⎛ x3 ⎞ r − 1 Let the constant term be ⎜ ⎟ ⎜ ⎟ . ⎜ r − 1⎟ ⎜⎝ x2 ⎟⎠ ⎝ 2 ⎠ ⎝ ⎠ ⎛ 10 ⎞ 11 − r 5 r −25 x ⎜ r − 1⎟ 2 ⎝ ⎠ This becomes after simplification. 2 r −1 If this is the constant term then the power of x must be zero. This gives us 5r – 25 = 0 and

r = 5. In other words, the constant term is the fifth term. ⎛ 10 ⎞ ⎜ ⎟ 211−5 x5⋅5−25 ⎜ 5 − 1⎟ 26 ⋅ 210 ⎝ ⎠ = = 840, which is Substituting r = 5 in the expression gives us 5−1 24 2 the constant term. 462

Algebra 10

EXAMPLE

98

Solution

Evaluate ⎛⎜ 5 ⎞⎟ + 2 ⎛⎜ 5 ⎞⎟ + 2 2 ⎛⎜ 5 ⎞⎟ + 2 3 ⎛⎜ 5 ⎞⎟ + 2 4 ⎛⎜ 5 ⎞⎟ + 2 5 ⎛⎜ 5 ⎞⎟. ⎝0⎠ ⎝1 ⎠ ⎝2⎠ ⎝3⎠ ⎝4 ⎠ ⎝5 ⎠ n n n Notice that the expression is the same as ( x + y)n = ⎛⎜ ⎞⎟ xn + ⎛⎜ ⎞⎟ xn −1y +...+ ⎛⎜ ⎞⎟ yn ⎝0⎠ ⎝1⎠ ⎝n⎠ for n = 5. ⎛5⎞ Since there is nothing but the coefficient ⎜ ⎟ in the first term, the first term in the ⎝0⎠ ⎛ 5⎞ binomial is 1. The term with the coefficient ⎜ ⎟ is 25. ⎝ 5⎠ We can rewrite the expression as follows: ⎛ 5 ⎞ 5 ⎛ 5 ⎞ 4 1 ⎛ 5 ⎞ 3 2 ⎛5 ⎞ 2 3 ⎛5 ⎞ ⎛5 ⎞ 5 4 ⎜ 0 ⎟ ⋅ 1 + ⎜ 1 ⎟ ⋅ 1 ⋅ 2 + ⎜ 2 ⎟ ⋅1 ⋅ 2 + ⎜ 3 ⎟ ⋅1 ⋅ 2 + ⎜ 4 ⎟ ⋅1 ⋅ 2 + ⎜ 5 ⎟ ⋅2 . ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

By the formula we have (1 + 2)5 = 35 = 243.

EXAMPLE

99

Solution

A rational term is a term that contains no irrational expressions.

7 What is the rational term in the expansion of ( 3 x + 2 x ) ?

Let us write the rational term as ⎛⎜ 7 ⎞⎟ ( 3 x )7− r+1(2 x )r −1 and try to solve for r. ⎝ r − 1⎠ ⎛ 7 ⎞ r −1 Simplifying gives us ⎜ ⎟2 x ⎝ r − 1⎠ make this rational.

In ⎛⎜ 7 ⎞⎟ 2 r −1 x ⎝ r − 1⎠ 13+r 6

However, x

rational term.

13+r 6

8−r 3

x

r −1 2

13+r 6

⎛ 7 ⎞ r −1 =⎜ ⎟2 x ⎝ r − 1⎠

. We need to find a value of r to

⎛ 7 ⎞ and 2r–1 have rational values for any integer r. , the factors ⎜ ⎟ ⎝ r − 1⎠ 13+5 6

is only rational for the value r = 5, so ⎛⎜ 7 ⎞⎟ 25−1 x ⎝ 5 − 1⎠

= 560 x3 is the

Check Yourself 12 1. What is the coefficient of the fourth term in the expansion of (x+b)15 ? 2. What is the seventh term in the expansion of (3x+y)11? 9

2 ⎞ ⎛ 3. What is the constant term in the expansion of ⎜ x4 + 2 ⎟ ? x ⎠ ⎝ Answers

⎛ 15 ⎞ 1. ⎜ ⎟ = 455 ⎜3⎟ ⎝ ⎠ Probability

2. 112266x5y5

3. 5376 463

EXERCISES

6 .4

A. Pascal’s Triangle and Binomial Expansion

⎛ 3

1. Expand each expression. a. (3x + 5)5

b. (2x2 – 5)4

B. Finding Binomial Terms Using Combination

c. (4x3 – 3y2)5

⎞

5

7. ⎜ 3 + 2 x ⎟ =...+( a ⋅ x)+... is given. Find a. ⎝x ⎠

2. What is the constant term in the expansion of (2x – 3)5?

8. Evaluate 3. Find the middle term in the expansion of

⎛9⎞ ⎛9⎞ ⎛9 ⎞ ⎛9 ⎞ ⎛9 ⎞ ⎜ ⎟ + 4 ⎜ ⎟ + 42 ⎜ ⎟ + 43 ⎜ ⎟ +...+ 4 9 ⎜ ⎟. ⎜0⎟ ⎜1 ⎟ ⎜2 ⎟ ⎜3⎟ ⎜9 ⎟ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

(2x – 5y)4.

4. Find the sum of the coefficients in each expansion. a. (3x + 2)4

b. (3x2 + 2y)15

9. What is the constant term in the expansion of 10

⎛ 3 1 ⎞ ⎜ 2x + 2 ⎟ ? x ⎠ ⎝

5. Find the twelfth term in the expansion of (3x + 1)15 when the terms are written in order of decreasing powers of x.

6. Find the sixth term in the expansion of (3x4 + y3)8

10. The sum of the coefficients in the expansion of

when the terms are written in order of decreasing powers of x.

(x + y)n is 1024. What is the greatest coefficient in this expansion?

464

Algebra 10

In this section we will learn some rules of probability which are frequently used for solving problems. rules of probability

Rule

1. 2. 3. 4. 5.

For every event E, 0 ≤ P(E) ≤ 1. For a sample space S, P(S) = 1. For two mutually exclusive events A and B, P(A ∪ B) = P(A) + P(B). For any two events A and B, P(A ∪ B) = P(A) + P(B) – P(A ∩ B). For any event A, P(A) +P(A′) = 1. In other words, P(A′) = 1 – P(A).

Note Many problems in probability are written in natural language. The key word for recognizing the union operation (∪) in a written problem is ‘or’. When we use the word ‘or’ (A or B) in mathematics, we mean A or B or both. The key word for recognizing the intersection operation (∩) in a written problem is ‘and’. When we use the word ‘and’ (A and B) in mathematics, we mean both A and B. EXAMPLE

100 A die is rolled. Find the probability that it shows 3 or 5.

Solution

Let T mean the die shows 3 and F mean the die shows S. Then ‘3 or 5’ means T ∪ F. Since T and F are mutually exclusive events, by the rules of probability we can write P(T ∪ F) = P(T) + P(F) 1 1 1 1 = + = . So the probability is . 6 6 3 3

EXAMPLE

101 A die is rolled. Find the probability that it shows an even number or a prime number.

Solution

The possible prime numbers are 2, 3 and 5 and the even numbers are 2, 4 and 6. Showing an even number (E) or a prime number (P) are not mutually exclusive events, since the outcome is in both events. 1 1 Since P(2) = , P( E ∩ P) = . 6 6 So the probability of E or P is P( E ∪ P ) = P( E) + P( P ) – P( E ∩ P )

1 1 1 = + – 2 2 6 5 = . 6 Probability

465

EXAMPLE

102 An urn contains five blue marbles, four red marbles and six yellow marbles. We want to take one marble from the urn. What is the probability of taking a red or a yellow marble?

Solution 1

Since a marble cannot be both red and yellow, drawing a red marble and drawing a yellow marble are mutually exclusive events. So the probability is 4 6 10 2 + = = . 15 15 15 3 We can also solve the problem in another way. Let E be the event that a red or yellow marble is drawn. Then the complement of E (written E′) is the event that neither a red nor a yellow marble is drawn. In other words, E′ is the event that a blue marble is drawn. P( R ∪ Y ) = P( R)+ P( Y ) =

Solution 2

P( E ') =1. We also know that P( E)+   

drawing a blue marble

So the probability of drawing a red or yellow marble is P( E) =1 − P( E ') =1 −

EXAMPLE

5 10 2 = = . 15 15 3

103 We have twenty cards numbered from 1 to 20. A card is drawn at random. What is the probability of drawing an even number or a number divisible by 3?

Solution

Let the event that an even number is drawn be E = {2, 4, 6, 8, 10, 12, 14, 16, 18, 20} and the event that a number divisible by 3 is drawn be T = {3, 6, 9, 12, 15, 18}. We can see that E ∩ T = {6, 12, 18}. So we can write P( E ∪ T ) = P( E)+ P( T ) − P( E ∩ T ) =

EXAMPLE

10 6 3 13 − = + . 20 20 20 20

104 A coin is tossed four times. What is the probability that the coin shows tails at least once?

Solution

The sample space contains 24 = 16 outcomes. If E is the event that we get tails at least once then E′ is the event that we get no tails. In other words, E′ is the event that we get heads three times (can you see 1 why?). Since there is only one way to do this, P( E ') = . 16 1 15 = So P( E) = 1 − P( E ') = 1 − . 16 16 We can check this result with the sample space: S = {HHHH, HHHT, HHTH, HTHH, THHH, HHTT, HTHT, THHT, TTHH, HTTH, THTH, HTTT, THTT, TTHT, TTTH, TTTT}.

466

Algebra 10

EXAMPLE

105 A

Solution

group of 6 people is selected at random. What is the probability that at least two of them have the same birthday? First let us assume that there are 365 days in a year. Then the sample space for one person’s birthday has 365 outcomes because there are 365 possible dates for a contains. Let the desired event be A. Then A′ is the event that none of these six people have a common birthday. So A′ contains 365 ⋅ 364 ⋅ 363 ⋅ 362 ⋅ 361 ⋅ 360 outcomes.

Let E be the sample space for the experiment. Then E contains 3656 possible outcomes, because there are six people. n( A ') 365 ⋅ 364 ⋅ 363 ⋅ 362 ⋅361 ⋅360 = 1− ≅ 0.05. So the probability of A is 1 − n( E) 3656

Check Yourself 13 1. A die is rolled. What is the probability that the die shows a number greater than 3 or an even number? 2. A number is drawn at random from the set A = {1, 2, 3, ... ,100}. a. What is the probability that the number is divisible by both 2 and 3? b. What is the probability that the number is divisible by 2 or 3? Answers 1. 2 2. a. 4 25 3

b. 67 100

A patient is talking to his doctor about a necessary operation. –– ‘I’m worried about this operation, Doctor. They say it’s 99 per cent risky.’ –– ‘That’s true, but you needn’t worry.’ –– ‘Why?’ –– ‘Because you are the hundredth patient. The other 99 have already died!’

Probability

467

In our probability studies up to now, we have considered sample spaces with only a small number of outcomes. These outcomes can be listed easily. But sometimes the sample space of an experiment has a large number of outcomes. Determining this number might not always be easy or practical. In such cases we can use the counting methods we learned in Chapter 1 to determine the number of outcomes in a sample space and an event.

EXAMPLE

is selected at random from the three-digit numbers formed using the digits 106 A{1,number 2, 3, 4, 5, 6, 7} without repetition. Find the probability that the selected number is an even number.

Solution

EXAMPLE

The number of outcomes in the sample space is 7 ⋅ 6 ⋅ 5 = 210. There are also three even numbers in the set {1, 2, 3, 4, 5, 6, 7}. So there are three possible choices for the units digit. There are then six possible choices for the hundreds digit and five possible choices for the tens digit. 90 3 = . So the number of outcomes in the event is 6 ⋅ 5 ⋅ 3 = 90 and the desired probability is 210 7

machine generates all the three-letter sequences of the letters in the word KAHVE, with 107 Arepetition allowed. Each sequence is written on a card and the cards are put in a box. Ömer draws a card. Find the probability that he draws a sequence beginning with the letter H.

Solution

EXAMPLE

There are 5 ⋅ 5 ⋅ 5 = 125 possible sequences, so n(S) = 125. If H is the event that the n( H ) 25 1 = = . sequence begins with H, n(H) = 1 ⋅ 5 ⋅ 5. So the probability is P( H ) = n(S) 125 5

108 Six people including Murat and Saim are to be seated around a circular table. Find the probability that Murat and Saim are seated next to each other.

Solution 1

We can use the formulas for circular permutation. The number of outcomes in the desired event

Solution 2

is (5 – 1)! ⋅ 2! = 4! ⋅ 2! and the number of outcomes in the sample space is (6 – 1)! = 5!. So the probability is 4!⋅ 2! = 2 . 5! 5 Alternatively, let us seat Murat first. Then we need to seat five more people in the remaining chairs, as shown opposite. Saim

Murat

can sit in any one of these chairs, so there are five possible places for him. But only two of the chairs are next to Murat, so 2 the probability that he sits in one of these chairs is . 5 468

Algebra 10

EXAMPLE

109 A box contains 4 yellow marbles and 6 red marbles. Two marbles are drawn at random from the box. What is the probability that both marbles are yellow?

Solution

We can draw any two marbles from ten marbles ⎛ 10 ⎞ without restriction in ⎜ ⎟ ways. ⎜2 ⎟ ⎝ ⎠ Similarly, we can draw two yellow marbles from ⎛4⎞ four yellow marbles in ⎜ ⎟ ways. ⎜2⎟ ⎝ ⎠ ⎛4⎞ ⎜ ⎟ ⎜2⎟ 2 ⎝ ⎠ So the desired probability is = . ⎛ 10 ⎞ 15 ⎜ ⎟ ⎜2 ⎟ ⎝ ⎠

EXAMPLE

110 A box contains 18 light bulbs. 5 of these bulbs are defective. We choose 3 bulbs at random. What is the probability that a. two of the chosen light bulbs are defective? b. at least one of the chosen light bulbs is defective?

Solution

a. We need to choose 2 defective bulbs from 5 bulbs and one working bulb from 13 bulbs.

⎛ 5 ⎞ ⎛ 13 ⎞ So the number of outcomes in the event is ⎜ ⎟ ⋅ ⎜ ⎟ =130. ⎝2⎠ ⎝ 1 ⎠ The number of outcomes in the sample space for selecting 3 ⎛ 18 ⎞ bulbs from 18 is ⎜ ⎟ =186. ⎝3⎠ ⎛ 5 ⎞ ⎛ 13 ⎞ ⎜ ⎟⋅⎜ ⎟ ⎝ 2 ⎠ ⎝ 1 ⎠ = 65 . So the probability is 408 ⎛ 18 ⎞ ⎜ ⎟ 3 ⎝ ⎠

b. Let the event be E. Then E′ is the event that no defective bulbs are selected. In other words E′ is the event that three working bulbs are selected. ⎛ 13 ⎞ ⎜ ⎟ 3 265 So the answer is P( E) =1 − P( E ') =1 − ⎝ ⎠ = . 18 ⎛ ⎞ 408 ⎜ ⎟ ⎝3⎠ Probability

469

EXAMPLE

111 A coin is tossed eight times. What is the probability of getting 5 heads and 3 tails?

Solution 1

The number of outcomes in the sample space is n(S) = 28 = 256. The number of outcomes in the desired event is n(E) = C(8, 5) ⋅ C(3, 3). ⎛ 8 ⎞ ⎛ 3⎞ ⎜ ⎟⎜ ⎟ n( E) ⎝ 5 ⎠ ⎝ 3 ⎠ 7 Therefore the probability is P( E) = = = . n(S) 256 32

Solution 2

Alternatively, we can think of the desired event as an arrangement of the letters in HHHHHTTT. By permutation with repetition, n( E) = So P( E) =

EXAMPLE

8! 8 ⋅7 ⋅6 = = 56. 5! ⋅ 3! 3 ⋅ 2 ⋅1

56 7 = . 256 32

112 In a group of 13 people, 4 people speak English, 6 people speak Turkish and 3 people speak German. A committee of 5 people is chosen at random from the group. What is the probability that the committee contains 2 English speakers, 2 Turkish speakers and one German speaker?

Solution

We can choose 5 people from 13 in C(13, 5) different ways without any restriction. So the number of outcomes in the sample space is C(13, 5). However, the desired selection can be achieved in C(4, 2) ⋅ C(6, 2) ⋅ C(3, 1) different ways. ⎛ 4⎞ ⎛6 ⎞ ⎛3⎞ 4! 6! 3! ⎜ ⎟⎜ ⎟⎜ ⎟ ⋅ ⋅ ⎜ 2 ⎟ ⎜ 2 ⎟ ⎜1 ⎟ n( E) ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ 2!⋅ 2! 2! ⋅ 4! 2! 30 . = = = So the probability is P( E) = 13! n(S) 143 ⎛ 13 ⎞ ⎜ ⎟ 5!⋅ 8! ⎜5⎟ ⎝ ⎠

EXAMPLE

student is taking a test which has 15 true-or-false 113 Aquestions. If the student guesses every answer, what is the probability that he or she will answer exactly eleven questions correctly?

Solution

There are two possible answers (true or false) for each question. Therefore the sample space has 215 outcomes ⎛ 15 ⎞ and the desired event has ⎜ ⎟ outcomes because the ⎝ 11 ⎠ event is the same as selecting 11 questions among 15. ⎛ 15 ⎞ ⎜ ⎟ 1365 ⎝ 11 ⎠ , So the probability is P( E) = 15 = 2 32768 or approximately 0.04.

470

Algebra 10

EXAMPLE

114 A machine generates a random four-letter sequence of letters from the letters in the word KARTAL. What is the probability that the word begins and ends with A?

Solution

Let us find the sample space. We have three cases: Case 1: 4! possible sequences do not contain an A, since there are four other letters to choose from {K, R, T, L}. ⎛4⎞ Case 2: ⎜ ⎟ ⋅ 4! possible sequences contain only one A. ⎝3⎠ ⎛4⎞ 4! Case 3: ⎜ ⎟ ⋅ possible sequences contain two A’s. ⎝ 2 ⎠ 2!⋅ 1!⋅ 1! ⎛4⎞ ⎛4⎞ 4! . So the sample space is 4 + ⎜ ⎟ ⋅ 4!+ ⎜ ⎟ ⋅ 3 2 2! 1! ⋅ 1! ⋅ ⎝ ⎠ ⎝ ⎠ ⎛4⎞ The number of outcomes of the desired event is ⎜ ⎟ ⋅ 2! because there are four letters left to ⎝2⎠ choose from if the two A’s have to be chosen. ⎛ 4⎞ ⎜ ⎟ ⋅ 2! 1 ⎝ 2⎠ = . So the required probability is 16 ⎛ 4⎞ ⎛4⎞ 4! 4! + ⋅ 4!+ ⎜ ⎟ ⋅  ⎜⎝ 3 ⎟⎠ 2 2! 1! 1! ⋅ ⋅ ⎠  sequences   

⎝

with no A sequences with one A

sequences with two A's

Check Yourself 14 1. A student will choose 4 courses at random to study next term. There are 14 courses in the list. Six of them are science courses. What is the probability that the student chooses all science courses? 2. A machine generates a three-letter sequence of letters from the elements in the set {a, b, c, d, e}, with repetition. What is the probability that the letters in the word are all different? 3. Ahmet, Kemal and their seven friends are called randomly to sit in 9 chairs placed side by side. What is the probability that Ahmet and Kemal are seated next to each other? 4. Set A has 6 elements. Each subset of A is written on a card and all the cards are put in a box. A student chooses a card at random. What is the probability that he selects a card which shows four elements? Answers 1. Probability

15 1001

2.

12 25

3.

2 9

4.

15 64 471

A STICKY PROBLEM! A box contains seven sticks which are respectively 2 cm, 3 cm, 4 cm, 5 cm, 7 cm, 8 cm and 11 cm long.

Find the probability that any three sticks chosen at random from the box will form a triangle.

Two mathematicians were talking about how important their jobs were. ‘My dear friend, our country is not taking math seriously. I think the government should tax people who can’t do math,’ complained one mathematician. ‘That is what the lotto is for!’ said the other. 472

Algebra 10

Many countries hold a 6/49 lottery. ‘6/49’ means that you must correctly guess six numbers from the first 49 positive integers to win the first prize. Does it sound easy? Only six numbers! In fact, it is not very easy to win a lottery like this. There are so many possible combinations of six numbers that the chances of choosing the right combination is very small indeed. Here are two sets of six lottery numbers: A = {1, 2, 3, 4, 5, 6} and B = {5, 12, 18, 23, 33, 41}. Which set of numbers do you think is more likely to win? Some people think that set B is more likely than set A. In fact, both sets of numbers have an equal chance, since six numbers are chosen from 49 at random. So what is the real probability that you will win the lottery with one ticket? The answer is

1 . In other words, about one in 14 million! Can you believe 13983816

it? Think about the math:

⎛ 49 ⎞ ⎟. ⎝6⎠

The number of different ways to choose six numbers from 49 is ⎜ This is equal to

49! 49 ⋅ 48 ⋅ 47 ⋅ 46 ⋅ 45 ⋅ 44 = = 13983816. Your (49 − 6)! ⋅ 6! 6 ⋅ 5 ⋅ 4 ⋅ 3 ⋅ 2 ⋅1

lottery ticket is only one of these sets of six numbers, so the probability that you will win the lottery with one ticket is

1 ≅ 0.00000007. 13983816

To be sure of winning the lottery, you would therefore need to fill in 13983816 tickets, using each possible combination of numbers just once. If it takes you 15 seconds to fill in one ticket, you would need approximately 58265 hours to complete them all. This is the same as 2427 days, or 6.65 years with no break. And of course, you would have to pay for all the tickets. Do you still want to play the lottery?

EXERCISES

6 .5

1. Ahmet, Berk, Can, Deniz and Engin are seated at

9. 5 boys and 4 girls are seated at random in a row

random around a circular table. What is the probability that Ahmet and Can are seated next to each other?

of 9 seats. What is the probability that the boys and girls are seated alternately?

2. Two integers are randomly selected from the set {1, 2, 3, 4, 5, 6, 7, 8, 9} without repetition. What is the probability that the sum of the numbers is 10?

3. Five math books and 4 physics books are arranged randomly on a bookshelf. What is the probability that books about the same subject are all together?

4. An integer is randomly selected from the set {1, 2, 3, 4, 5, 6, 7, 8, 9} and then replaced. Then another integer is selected. What is the probability that the sum of the two integers is 10?

5. Nine people want to be seated around a circular table. What is the probability that two particular people will not be seated next to each other?

10. Two dice are rolled together 900 times. How many times would you expect to get a sum of 7?

11. There are 4 different letters and 6 different mailboxes. Each letter will be put in a random mailbox. What is the probability that all four letters will be put in the same mailbox?

12. 200 patients were treated with a new medicine. 32 of them were not cured. Four patients from the 200 are selected at random. What is the probability that three of them were cured?

13. The first 5 positive integers are written randomly in a row. Find the probability that the numbers are written in either ascending or descending order.

14. A four-digit number is formed using the digits 6. Two numbers are drawn at random from 5 positive and 3 negative numbers. What is the probability that the product of the selected numbers is positive?

{1, 2, 3, 4, 5, 6, 7} without repeating the digits. Find the probability that the number is an even number.

15. A box contains 4 red marbles and 8 other marbles 7. Two numbers are selected randomly from 6 odd numbers and 5 even numbers. What is the probability that the product of the selected numbers is even?

of different colors. What is the probability that 4 marbles selected at random from the box are not red?

16. Seven points A, ..., G are arranged in a circle. A 8. In a lottery game, a player must pick 4 winning numbers from 42 numbers to win the prize. Find the probability of winning the prize. 474

triangle is drawn by connecting three points chosen at random. What is the probability that point E is a vertex of the triangle? Algebra 10

17. Three hundred people apply for 4 jobs. Ninety of

24. A committee of 4 men and 3 women is chosen at

the applicants are women. Four people are selected at random for the jobs. Find the probability of each event.

› random from a group of 8 men and 5 women that

a. all the selected people are men

25. A teacher distributes 20 questions before an exam

b. exactly two people are men

includes Sami and Dilara. What is the probability that both Sami and Dilara are chosen?

› and tells her students that 10 of them will be in

exam. Mehmet can solve 15 of the questions. In order to pass the exam a student must answer at least 6 questions correctly. What is the probability that Mehmet will pass the exam?

c. exactly one person is a man d. no men are selected

18. A standard piano keyboard has 88 different keys. A cat jumps on 6 keys of the keyboard at random (possibly with repetition). Find the probability that the cat will strike the first six notes of Beethoven’s Fifth Symphony.

19. The square in the figure is divided into 9 smaller squares. A quadrilateral is chosen at random from all the quadrilaterals in the figure. What is the probability that the quadrilateral is a square?

20. An urn contains 3 red, 3 blue and 4 yellow

26. Abraham and Bill are in a group of 12 people.

› Seven people are chosen randomly from the group

and seated in a row. Find the probability that Abraham and Bill are seated next to each other.

27. Three numbers are selected from the set

›› {1, 2, 3, ..., 15}. What is the probability that the

sum of the selected numbers is divisible by 3?

28. We draw two numbers at random from the set

›› {1, 2, 3, ..., 100}. What is the probability of

drawing a pair of consecutive numbers?

29. A committee of 4 people is chosen from 6 men and 5 women. What is the probability that the committee contains at least one man?

marbles. Three marbles are drawn at random. Find the probability no red marbles are drawn.

21. A box contains 4 red, 5 yellow and 4 white marbles. ›

We draw 3 marbles at the same time. What is the probability that only one of them is a red marble?

22. Three numbers are selected from the set ›

{1, 2, 3, ..., 15}. What is the probability that the product of the numbers is divisible by 3?

30. A group of 4 people is chosen at random from 6

›› couples. What is the probability that there is no

couple in the group?

31 . A chessboard has 64 squares and the length of the

›› side of each square is 1 unit. A rectangle is

drawn at random on the chessboard. What is the probability that the perimeter of the rectangle is greater than 4 units?

23. A subset is drawn from all of the four-element ›

subsets of the set A = {a, b, c, d, e, f, g, h}. What is the probability that the selected subset does not contain the element d?

Probability

32 . Two numbers are drawn at random from the set {1, 2, 3, ..., 100}. Find the probability that one of the drawn numbers is half of the other.

››

475

CHAPTER REVIEW TEST

6A 5. A box contains 5 different green balls and 9

1. There are 3 routes from city A to city A B C B, 4 routes from city B to city C and 2 direct routes from A to C. In how many different ways can Serkan travel from A to C? A) 7

B) 9

C) 10

D) 12

E) 14

different blue balls. In how many different ways can Salim pick out a green and a blue ball in any order? A) 54

B) 45

C) 40

D) 35

E) 9! – 5!

6. In how many ways can we name the vertices of a pentagon using any five of the letters O, P, Q, R, S, T, U in any order?

2. In each sequence below, L represents one of the

A) 2520

letters in the 26-letter English alphabet and D represents a digit. A factory wants to register 10 million different product items. Which sequence shows a possible registry format? A) LLDDD

B) LDDL

D) DDDLLL

C) DDDDL

B) 9040 D) 4880

C) 5140

E) 3600

⎛1 2 3 4 ⎞ ⎟ and ⎜1 3 4 2 ⎟ ⎝ ⎠

7. The permutation functions f = ⎜

⎛1 2 3 4 ⎞ ⎟ are given. Find the permutation g=⎜ ⎜4 2 3 1 ⎟ ⎝ ⎠

E) DDLLL

function h such that f D g = h.

3. How many two-digit odd numbers can be formed from the digits {1, 2, 3, 4, 5, 6, 7} if repeated digits are allowed? A) 14

B) 42

C) 28

D) 21

E) 49

⎛1 2 3 4 ⎞ A) h = ⎜ ⎟ ⎜3 4 2 1 ⎟ ⎝ ⎠

⎛1 2 3 4 ⎞ B) h = ⎜ ⎟ ⎜2 3 4 1 ⎟ ⎝ ⎠

⎛1 2 3 4 ⎞ C) h = ⎜ ⎟ ⎜3 4 1 2⎟ ⎝ ⎠

⎛1 2 3 4 ⎞ D) h = ⎜ ⎟ ⎜2 3 1 4⎟ ⎝ ⎠

⎛1 2 3 4 ⎞ E) h = ⎜ ⎟ ⎜4 2 1 3⎟ ⎝ ⎠

4. Evaluate

( n + 2)!( n − 2)! . ( n +1)!( n − 1)!

A) (n – 3)

B) (n – 1) D)

476

( n + 2) ( n − 1)

8. How many six-digit numbers can be formed from C) E) 49

( n +1) ( n + 2)

the digits {2, 3, 4, 6, 7, 8} without repetition if the digits 3 and 7 must be together? A) 120

B) 180

C) 144

D) 96

E) 240

Algebra 10

9. How many eight-letter sequences can be formed from the letters in the word ALTAYLAR? A) 2120

B) 2480 D) 3360

C) 3200

E) 3640

13. There are 11 students of different heights in a class. We choose any 4 students and line them up from tallest to shortest in order. How many different orders are possible? A) 280

B) 330

C) 480

D) 660

E) 7920

14. What is the middle term in the expansion of (2x + 5y)4? A) 600x2y2

D) 6x2y2

10. In how many different ways can 5 couples be seated around a circular table if the couples must not be separated? A) 768

B) 724

C) 844

D) 696

E) 576

11. Which one of the following is a mathematical combination?

B) 120xy2

C) 5000xy3 E) 160x2y2

15. There are 8 different math books, 4 different biology books and 6 different geometry books on a table. Nuran wants to select 4 math books, 3 geometry books and 2 biology books and then arrange them on a bookshelf so that books on the same subject are together. How many different arrangements are possible? ⎛ 8 ⎞⎛ 4 ⎞⎛ 6 ⎞ A) ⎜ ⎟⎜ ⎟⎜ ⎟ 9! ⋅ 3! ⋅ 4! ⋅ 2! ⎝ 4 ⎠⎝ 2 ⎠⎝ 3 ⎠ ⎛ 8 ⎞⎛ 4 ⎞⎛ 6 ⎞ B) ⎜ ⎟⎜ ⎟⎜ ⎟ 3! ⋅ 3! ⋅ 4! ⋅ 2! ⎝ 4 ⎠⎝ 2 ⎠⎝ 3 ⎠

A) a social security number B) the key for a combination lock C) a committee chosen from a group of 10 people

8 4 ⎞⎛ 6 ⎞ C) ⎛⎜ ⎞⎛ ⎟⎜ ⎟⎜ ⎟ 3! ⎝ 4 ⎠⎝ 2 ⎠⎝ 3 ⎠ 8 4 ⎞⎛ 6 ⎞ D) ⎛⎜ ⎞⎛ ⎟⎜ ⎟⎜ ⎟ 9! ⎝ 4 ⎠⎝ 2 ⎠⎝ 3 ⎠

D) the PIN code for a cellular phone E) your name and surname

E) 3! ⋅ 4! ⋅ 2!

16. A company in Ankara will send a group of 10

blue balls in a box. In how many different ways can Salih pick out two balls?

managers to participate in a seminar in Istanbul. However, only 6 plane tickets are available. The rest of the managers will go by bus. Two people are afraid of flying so they do not want to go by plane. In how many ways can the group be divided for the journey to Istanbul?

A) 132

A) 18

12. There are 5 different green balls and 9 different

B) 124

Chapter Review Test 6A

C) 111

D) 104

E) 91

B) 24

C) 28

D) 42

E) 56 477

CHAPTER REVIEW TEST

6B 6. Fatma has 5 different history books and 4

1.

different math books. In how many ways can she arrange them all on a shelf so that the history books are together and all the books are between two math books? A) 2640

B) 2880

D) 8640

C) 5160

E) 12520

In how many different ways can the mouse in the picture get to the cheese without passing by a cat if all the gates allow only one direction pass? A) 48

B) 36

C) 24

D) 44

E) 60

7. In how many ways can we exhibit 5 of 8 different new cars in a row if a certain car must be the first on the right?

2. What is the sum of all the four-digit numbers which can be made from the elements of the set {1, 2, 3, 4} with no repeated digits? A) 66660

B) 62600

D) 48800

A) 720

B) 760

C) 840

D) 900

E) 960

C) 57420 E) 36000

3. How many three-digit even numbers smaller than

8.

550 can be formed from the digits in the set {0, 3, 4, 5, 6, 7, 8} if no digit can be used more than once in a number? A) 24

4. Solve

B) 32

C) 35

D) 42

E) 45

(2 n)! n! ÷ = 28 for n. (2 n − 3)! ( n − 2)!

from the letters in the sequence THISWORD if no letter can be used more than once?

The cube in the figure is made up of 27 sticks. An ant is trained to walk up and right along the sticks, relative to the cube. If it starts at point A and walks only along the sticks, in how many different shortest ways can it get to piece of the sugar at point S?

A) 1680 B) 1540 C) 1420 D) 136 E) 1260

A) 90

A) 2

B) 3

C) 4

D) 5

E) 6

5. How many four-letter sequences can be formed

478

B) 120

C) 144

D) 150

E) 180 Algebra 10

9. How many seven-digit odd numbers can be formed

14. What is the coefficient of the term containing

by rearranging the digits in the number 5321233?

x12y6 in the expansion of (x3 – 2y2)7?

A) 144

A) 84

B) 168

C) 196

D) 225

E) 240

B) –280

C) 560

D) 448

E) 35

10. How many different necklaces can we make by threading 3 different red beads, 3 different green beads and 3 different yellow beads onto a chain if beads of the same color must be kept together? A) 240

B) 216

C) 196

D) 164

E) 144

11. The figure shows ten points. Three of the points are chosen at random to form a triangle. How many different triangles can be constructed? A) 80

B) 90

C) 100

D) 110

15. A group of 5 people will be selected from 4 doctors and 7 nurses such that the group contains at least one doctor. After the selection, the group will hold a meeting around a circular table. How many different seating arrangements are possible around the table? 4 7⎞ A) ⎛⎜ ⎞⎛ ⎟⎜ ⎟ 4! 1 ⎝ ⎠⎝ 4 ⎠

11 B) ⎛⎜ ⎞⎟ 4! ⎝5⎠

C) ⎛⎜ 7 ⎞⎟ 4! ⎝ 5⎠

⎡ 11 7 ⎤ D) ⎢⎛⎜ ⎞⎟ − ⎛⎜ ⎞⎟ ⎥ 4! 5 ⎣⎝ ⎠ ⎝ 5 ⎠ ⎦ 7⎞ E) ⎛⎜ 4 ⎞⎛ ⎟⎜ ⎟ 4! ⎝ 2 ⎠⎝ 3 ⎠

E) 120

12. Six women and 8 men are in a table tennis club. If each game needs 2 players for each team, how many different games between men and women can be arranged? A) 240

B) 276

C) 360

D) 420

E) 480

13. A committee of 4 people will be selected from 8 girls and 12 boys in a class. How many different selections are possible if at least one boy must be selected? A) 2865 D) 4455 Chapter Review Test 6B

B) 3755

C) 4225 E) 4775

16. A group of 12 people booked tickets to the theater. When they arrive they find that 5 places in the front row and 7 places in the back row have been reserved for them. 3 people do not want to sit at the front and 2 of them do not want to sit at the back. In how many different ways can the 12 people be seated? ⎛ 10 ⎞⎛ 7 ⎞ A) ⎜ ⎟⎜ ⎟ ⎝ 3 ⎠⎝ 4 ⎠

B) 5! ⋅ 1!

C) ⎛⎜ 7 ⎞⎟ 12! ⎝ 3⎠

9⎞ D) ⎛⎜ 12 ⎞⎛ ⎟⎜ 4 ⎟ 3 ⎝ ⎠⎝ ⎠

E) ⎛⎜ 7 ⎞⎟ 5! ⋅ 7! ⎝ 3⎠ 479

EXERCISES 1. a. f: b.

→

b. g:

– [–1, 4) c.

1 .1 – {5} →

c. h:

3. a. [–11, 7] b. (–3, 5]

– (0, 5]

6. a. g(x) = x + 1, h(x) = ñx, t(x) = b. f –1:

– {2} →

– {3}, f –1(x) =

3 9. a = , b = 4 2

– {0} →

2. a. [–ò13, ò13]

– {3} d. t: (–4, ∞) – {1} → (–2, 3] c. [5, 8)

4. a. [0, 2] b. [0, 4]

1 , r(x) = 5 – x, f(x) = r(t(h(g(x)))) 7. a. f –1: x

5. a. (ñx + 1)2 b. →

–3x – 3 x – 1+ 2 c. f –1: [1, +∞) → [1, +∞), f –1(x) = x+ 2 2

11. a. (– ∞,

10. a. increasing b. increasing c. decreasing

7 ] 4

x2 +1

, f –1(x) = 2 – 5x 8. a.

7 b. [ , ∞ ) 4

1 b. 2 2

12. a. odd b. even

c. odd d. odd e. odd

EXERCISES 1. 2

2. a.

1 .2 y = f(x)

1 O

y

b.

y

1

x

6

y = f(x)

y = f(x)

1

4

–1

–2 O –1

1

O –1

x

x

⎧⎪ x + 3 if x ≥ –3 3. a. f ( x) = ⎨ ⎪⎩ – x – 3 if x < –3

⎧⎪ 2 x if x ≥ 0 b. f ( x) = ⎨ ⎪⎩ 0 if x < 0

x<2 ⎧ –2 x +5 if ⎪⎪ d. f ( x) = ⎨ 1 if 2 ≤ x < 3 ⎪ x>3 ⎪⎩ 2 x – 5 if

⎧ –2 x ⎪⎪ e. f ( x) = ⎨ –2 ⎪ ⎪⎩ 2 x

480

y

c.

if

⎧⎪ x2 – x – 2 if x ∉ (–1, 2) c. f ( x) = ⎨ ⎪⎩ – x2 + x + 2 if x ∈ (–1, 2) x < –1

if –1 ≤ x < 1 if

x ≥1

⎧⎪2 x2 + 2 x if f. f ( x) = ⎨ if ⎪⎩ –6 x

x ≥ –4 x < –4

Answers to Exercises

4. a.

b.

y

c.

y

y = |–x| 2

y

e.

x

1 21

–1 O

x

b. x < 1

x

5

c. x ∈ {–1, 0, 1}

x

O

–2

–1 O

d. x ∈ {–1, 0, 1}

3 y = |5x + 4| + 2x – 1

O

x

5. a. x > 0

–4 5

y = |x2 – 4x – 5|

5 1

y

f.

3

–3 –2 –1

9

1

O

y = x|x + 1| + 3

y

y = |x2 – 1|

y = |2 – 4x|

x

O

d.

y

–2 –3

6. a.

b.

y y = sgn(x2)

1

y

y = sgn(x – 1)

1

x

O

c.

y

x

1

O –1

1 –1 O

y

b.

y = f(x)

2 –2 –11 O1 2 –1

y

y = f(x)

c. 5 4 3 2

x

5 7 ) e. x ∈ (– , – 1] 2 3

x

y 4

1

1 2

–1 – 1 2 O

1

d.

y y = f(x)

1 –3 –2 –1 O

x

2

–1

7. a. x ∈ [–2, –1) b. x ∈ [2, 3) c. x ∈ [–3, –2) d. x ∈ [1, 8. a.

1 y = sgn(xx + – 2)

1

x

y = f(x)

2 1

–1

–2 –1 O

1 2

x

e.

y

–2 –1

–2 –3

⎧ x2 + 2 x if x > 1 ⎪⎪ 9. x ∈ {–1, 4} 10. f ( g( x)) = ⎨ 3x + 4 if 0 < x ≤ 1 ⎪ ⎪⎩ 3x +7 if x ≤ 0

⎧⎪ x – 1 if x > 1 12. a. f ( x) = ⎨ ⎪⎩ 1 – x if x ≤ 1

Answers to Exercises

11.

f –1 :

y = f(x)

4 3 2 1 O1 2 –1 –2 –3 –4

x

⎧ x +1 if x > 3 ⎪ 4 ⎪ → , f –1( x) = ⎨ ⎪ x–1 if x ≤ 3 ⎪⎩ 2

⎧⎪ x2 – 3x – 4 if x ∉ (0, 3) b. f ( x) = ⎨ ⎪⎩ – x2 + 3x – 4 if x ∈ (0, 3)

⎧⎪ x3 – x if x ≥ 0 c. f ( x) = ⎨ 3 ⎩⎪ – x – x if x < 0

481

⎧⎪ x2 if x ≥ 0 13. a. f ( x) = ⎨ 2 ⎩⎪ – x if x < 0

⎧ –2 x +1 if x < 0 ⎪⎪ d. f ( x) = ⎨ 1 if 0 ≤ x < 1 ⎪ ⎪⎩ 2 x – 1 if x ≥ 1

EXERCISES 1. 17 y

8.

2. 24

3. 27

y = f(x)

5

x<2 –1 if ⎧ ⎪⎪ e. f ( x) = ⎨ 2 x – 5 if 2 ≤ x < 3 ⎪ x≥3 1 if ⎪⎩

4. 17 y

–2

x

6. 2a3 + a2

5. 2k + 1

9.

5

3

⎧ ⎪ f. f ( x)= ⎪⎨ ⎪ ⎪ ⎩

x2 – x – 1

x ≤ –1

if

2

– x – x +1 if –1 < x < 0 0 ≤ x <1

– x2 + x +1 if 2

x +x –1

x ≥1

if

2 .1 7. b2 – 2b + 1

a. 0 b. 4 c. –4

y = f(x)

4 3

O

2x + 3 if x ∉ (–1, 3) ⎧⎪ c. f ( x) = ⎨ ⎪⎩ 2 x2 – 2 x – 3 if x ∈ (–1, 3)

⎧⎪ 2 x – 1 if x ≥ 1 b. f ( x) = ⎨ if x < 1 ⎪⎩ 1

y

10.

1

–5

c. 2 x

O

–3

x

O 2

a. –2 b. does not exist

2 y = f(x)

–2

–4 –7

11. (11.7, 12.3)

–5 –5.1

–4.9

13. (3.99, 4.01)

17. p ∈ {–2, –1, 0, 1, 3} 18. a. lim+ f ( x) = 0,

e. 1 f. 2

lim f ( x) = 0

b. x = –1, x = 1

22. 28

x→1

23. a. 1 b. 0 c. 1 d. –1 e. –1 f. –1

c. 1

d. –2

e. –2

f. 2

31. –6

16. a. –3 b. 1 c. –1 d. 0 lim f ( x) =1, lim+ f ( x) =1,

x →0–

x →1

20. x = 0

21. a. does not exist

15 e. –11 f. 0 8

26. a. –2 b. 0 c. ∞ d. does not exist e. 1

28. a. ∞ b. ∞ c. –∞ d. does not exist e. –∞

f. –∞ g. ∞ h. 0 30. a. –∞ b. –∞ c. –∞ d. –∞

EXERCISES

b. x = –2

24. a. 5 b. –1 c. 0 d.

b. does not exist c. –1 d. –2 e. does not exist f. 0 g. 31 b. 1

x→0

19. a. lim f ( x) = –4, lim f ( x) = –3 x→0

14. (4.95, 5.05)

lim f ( x) =1, lim+ f ( x) =1,

x → –1–

x → –1

x →1–

b. 1

12. (–5.1, –4.9)

32. a.

25. a. 4 27. a.

2 3

29. a. 0 b. –∞ c. ∞ d. ∞ e. –∞

3 1 6 b. 16 c. 0 d. 2ñ3 e. 3 2

2 .2

1. a.

1 5 b. c. –7 d. 15 e. 6 f. 2 g. 1 h. m i. 16 10 3

2. a.

1 5 1 1 2 b. c. 3 d. e. f. g. –1 h. 1 i. 2 2 7 2 2 2

j. –6

3. a. 3 b. 0 c. –∞ d. –∞ e. 0 f. 1 g. –1 h. 2

4. a.

6 7

6. a.

b. 5 c. 3 d.

2 5

5. a. –

1 5 1 b. – c. 2 d. 7. a. e2 b. e–9 c. e3 d. e1/3 e. e–4/3 f. e6/5 g. e14/5 h. e–18 i. e–1 j. e4 2 2 2

482

4 9

b. –

1 1 c. –∞ d. 2 2

Answers to Exercises

EXERCISES

2 .3

1. continuous at x = –2, x = –1, x = 0, x = 2, discontinuous at x = –3, x = 1, x = 3 3. discontinuous c.

– {1} d.

4. x = 1, x = 2 – {–1, 6} e.

5. continuous

– {0} f.

c. {x | 2kπ ≤ x ≤ (2k + 1)π, k ∈ } d.

EXERCISES

g.

7. m = 0, n = ±2

6. 3

2. continuous

8. a ∈ {1, ±ñ2}

9. a.

b.

b. – {x | x = 2kπ, k ∈ } – [–1, 4] 10. a. 5 3 – [–2, 2] g. [– , – ] 11. – 6 12. 4 2 4

h.

– (–2, 2)

– {2} e. [3, 5] f.

3 .1

1. a. fourth b. third c. third d. third e. third f. second 2. a. 0.785 b. 1.832 c. –1.221 d. 1.346 e. 16.336 24π 3. a. 154° b. 359° c. 10 π d. 19 9

⎛ ⎞ 4. a. (–1, 0) b. ⎜ − 2 , − 2 ⎟ ⎜ 2 2 ⎟⎠ ⎝

5. 108°

6. a. x = 105° + (k ⋅ 180°)

34 2 π k ⋅ 2π d. x = + 7. a. ò17 b. c. 8. ò91 9. 6 10. 8+ 2 3 11. 13 6 6 3 3 729 9 12. a. b. 12 13. units 14. a. sec x b. 1 c. csc x ⋅ sec x d. sin x e. sin x f. cot x 15. a. 1– sin x b. 1– sin 32 4

b. x = π + 24kπ

c. x = k ⋅ 90°

x c. sin2 x d. cot x e. 1 f. (sec x + 1)2 18. a. 1 b. 3

EXERCISES 1. a. 0 e.

b. 0

c. –1

2 kπ − { + }, k ∈ 3 3

c. [–7, –1] b. –

c. +

3 .2 d. 3

e. 0

2. a.

b.

5 π kπ − { + }, k ∈ 18 3

f.

d. [–2, 6] e. [0, 6] d. + e. +

19. 2

f. –

d. ∞, –∞ e. ∞, –∞ f. ∞, –∞

f. [–4, 0] g. –

g.

g.

h. +

h.

h. i. –

1 π kπ − { + + }, k ∈ 2 8 4

c.

j. +

π kπ nπ − { + } ∪ { }, k, n ∈ 4 2 5

i. (–∞, 0] ∪ [2, ∞) k. +

l. –

− {–1 + kπ}, k ∈

d.

3. a. [–1, 1]

j. (–∞, –3] ∪ [1, ∞)

5. e, f, j, k

6. a. 7, 1

k.

b. 2, –2

7. a. cos 299° b. sin 66° c. tan 88° d. cot 30° 8. a. y < x < z

b. [–5, 5] 4. a. + c. 4, –6

b. b < a < c < d

c. r < m < n d. y < x < z < k e. b < a < c 9. a, b, c, d, f 10. a, d, e, f 11. a, b, d 13. a. 0 b. 1 14. a. 12° b. 68° c. 32° d. 48° e. 50° f. 30° g. 70° h. 10° 15. a. 25° b. 40° c. 65° d. 55° e. 5° f. 85° g. 80° h. 40° 16. a. 80° b. 36° c. 60° d. 30° e. 20° f. 51° g. 80° h. 5° 17. a.

π 11

π π g. 5π h. 2π 18. a. b. 3π c. d. 5π e. π f. π g. 5π h. 2π 8 12 15 12 19 14 6 9 17 10 π 2 1 1 15π 3 19π f. g. h. 21. a. b. − c. –1 d. –1 e. f. 7 2 2 2 73 2 99

c. ñ3

d. –ñ3

Answers to Exercises

e. − 1 2

f.

2 2

g. –1

h. − 3 3

23. a. 1 2

b. 1 2

b. 5π c. 5π d. 3π e. 2π f. 2π 12 13 5 13 14 π 19. a. b. π c. 3π d. 4π e. π 6 3 7 13 15 g. 1

h. –ñ3

c. 1 d. – ñ3 e. − 2 2

22. a.

2 2

b. − 3 2

f. − 1 g. –1 h. –ñ3 2 483

2 24. a. − 1 b. − 1 c. –1 d. ñ3 e. − f. − 1 g. 1 h. ñ3 25. a. ñ2 b. 0 c. 6 26. a. tan α + cot α – 2sin α b. 0 2 2 2 2 2 , tan θ = –1, cot θ = –1 c. sin θ = − 5 , cos θ = − 4 , cot θ = 4 27. a. cos θ = 1 , tan θ = 3 , cot θ = 3 b. sin θ = 2 2 3 5 41 41 3

3 2 28. a. cos θ = , tan θ = –1, cot θ = –1 2 13 13 2 7 7 24 4 1 1 , cot θ = , cos θ = − , cot θ = − b. sin θ = − , tan θ = c. sin θ = 25 24 7 4 17 17 2 3 17 7 2 1 1 1 7 3 cos θ = , tan θ = d. sin θ = 29. a. b. c. d. 4 30. a. cot α = b. cos α = − 10 13 10 7 3 13 4 25

d. sin θ = −

c. sec α =

2

, cos θ =

, tan θ = −

2

31. a. ò57 b. 3 3 32. P = 26, A = 21ñ3 33. a. 48° 30′ b. 136° 12′ c. 213°45′ d. 313° 47′ 34. a. 121.25° 3 2

– b. 346.8 3° c. 198.32° d. 23.94° 35. a. 0.3971 b. 0.5842 c. 2.9208 d. 0.4975 36. a. 0.8549 b. 0.9737 c. –1.1323 π π 2π 2π π 2π π π d. 0.0032 37. a. b. c. π d. e. f. 38. a. b. c. 5π d. 39. 2π2 40. a. 3 5 2 5 5 5 3 3 π b. 2π c. 2π d. 2π e. π f. π g. 2π h. 2π i. π j. 3π 41. π 42. 2 2 43. a.

y 3 –p 6

p 2

1 p 6

0

–p 3

–1

b. 5

p 3

2p 3

x

c.

y

y

3

1 –p

d.

p

p 2

–p 2

3p 2

2p

x

–

p 2

–

p 3

–

p 6

p 6

p 3

p 2

x

y –p

p p – 3p – – 4 2 4

p 4

p 2

3p 4 p

3p 2

2p

x –2 –4

484

Answers to Exercises

e.

44. a.

y

y 1

6

p 2

1 – p – –3p – p 2 2 4

p 8

p 4

p 8

3p 8

p 2

p 6

x

p 3

5p 6

2p 3

p

x

–1 –4

b.

c.

y p 4

d.

y

2

3p 4 p

p 2

ñ2 x

p

2

–1

p 2

p

x

x

p 2

e.

y

y 3 3 2 p

45. III, V, VI e. 0° b.

3 2

c. 2ñ2

46. a. y = 2 sin x

f. 150° c.

1 2

d. –

x

g. 90° d.

1 3

12 13

e.

h. 180° e. 0 f.

17 17

1⎤ ⎡ b. ⎢ −1, − ⎥ c. [2, 4] d. 3⎦ ⎣ 48+ 25 3 60. − 61. 0 11 Answers to Exercises

3 4

f. ñ3

c. y = 3+sin x

b. y = sin 3x 48. a.

π 4

50. a. 52. a.

⎡2 4⎤ ⎢5 , 5⎥ ⎣ ⎦

e.

62. 1

63.

4 7

b. – 3 2

π 3

b.

64.

c. 2 2

5 2

c.

54. −

π 4

b. –

f.

π 4

π 4

d.

c.

3 2

3 3

π 2

47. a. 0° e.

5π 6

f. –

d.

π 3

d. –

24 7

e.

56.

5π 6

55. 0

e. −

π 6

π 4

3 7 7

57.

b. –45° π 3 π f. − 4

g. –

f.

120 169

17 13 65

c. –90° h.

2π 3

d. 45° 49. a. 1

24 4 b. 25 5 3⎤ ⎡ 5 53. a. ⎢ − , − ⎥ 2 2 ⎣ ⎦ π 4 58. 59. – 3 5

51. a.

5 2 485

3 .3

EXERCISES

13 − 6 3 b. −3+ 37 c. 1 + ò13 d. 2 5 − 2 3 2. a. 2

1. a.

113 b. ò13 c. 3

14 d. 2

19 e. 2 f. 2

79 3. 120° 2

14 3 d. 2 7. π units 8. a. 5ñ6 cm b. 5ñ2 cm 3 3 17 9. 6 cm 10. m(∠A) = 30°, m(∠B) = 60°, m(∠C) = 90° 11. 84 cm2 12. units 13. a. 2 ( 3 +1) 2 4

10 7 3 − 15 4. 135° 5. a. ñ2 + ñ6 b. ñ2 c. 9.6 6. a. 5 b. c. 10 16 6

b.

2 2 2 1 5 3 (1 − 3) e. ñ3 – 2 f. 2 – ñ3 g. (1 − 3) h. –2 – ñ3 14. a. + ( 3 – 1) c. 2 + ñ3 d. 4 4 4 6 12

1 5 3 5⎛1 3 34 5 5 5⎛1 ⎞ ⎞ c. – ⎜ − 3 ⎟ d. ⎜ + 3 ⎟ e. − 55 (9 3 + 32) f. − 71 (9 3 + 32) 15. a. 1 b. 34 6 12 6 ⎝2 6 ⎝2 ⎠ ⎠ 2 47 17 4 17 c. d. e. 0 f. 16. a. cos x b. 2cos x cos y c. 3 sin x d. 2sin y cos x 2 289 17 3tan α − tan 3 α 24 7 24 18. a. sin 2 x = , cos 2 x = , tan 2 x= 17. cos 3α = 4cos3 α – 3cos α, tan 3α = 2 25 25 7 1 − 3tan α

b.

b. sin 2 x = –

120 119 120 , cos 2 x = – , tan 2 x= 169 169 119

d. sin 2 x = –

15 7 15 , cos 2 x = , tan 2 x= – 8 8 7

19. a. sin

5 x x 2 5 x 1 = , cos = , tan = 2 5 2 5 2 2

c. sin 2 x = – e. sin 2 x =

b. sin

336 527 336 , cos 2 x = , tan 2 x= – 625 625 527

12 5 12 , cos 2 x = – , tan 2 x= – 13 13 5

x 5 x 2 5 x 1 = , cos = – , tan = – 2 5 2 5 2 2

c. sin

1 2 2 +3 1 x x x = , cos = , tan = 2 2 2 2 2 +3 18+12 2 18+12 2

d. sin

1 x x = , cos = 2 2 4+ 2 2

f. sin

x 1 x 5+ 26 x 1 = , cos = − , tan = − 2 2 2 5+ 26 52+10 26 52+10 26

e.

2 f. g. 2 h. 0 2 2

e. −2 cos 26. −

8 11

35. − 6 5 486

2 sin

π 8

2 +1 4+ 2 2

, tan

x = 2 −1 e. sin x = 6 , cos x = – 30 , tan x = – 5 2 2 6 2 6 2 5

21. a. 2sin 4x ⋅ cos x b. −2 cos

20. a.

2 2

b.

6 2

c. −

2 2

d.

2 ( 3 − 1) 2

5x 3x 11x 7x ⋅ sin c. 2sin 5x ⋅ sin x d. 2 cos ⋅ cos 2 2 2 2

3x 9x 5x 7x x 33 f. 2 sin g. sin 11x + sin 9x = 2sin 10x ⋅ cos x h. 2 sin ⋅ sin x 24. 25 25. – ⋅ cos ⋅ sin 2 65 2 2 2 2 19

27.

3 10 10

36. 2ñ6

37.

28.

2 ( n +1)2

1 16

38.

29. 3 4 3 + 2x 4

30. −

33 56

39. − 3 4

2 31. sin 40 ° = 2 x 1 − x

40. − 2 4

32. −

55 24

33. –

3 4

34. 2–ñ3

41. 1 Answers to Exercises

3 .4

EXERCISES 1. a.

π 3π + 2 kπ, + 2 nπ, k, n ∈ 4 4

c. ( −1)k +1

π kπ + , k∈ 12 2

c.

11π kπ + , k∈ 60 5

b.

π kπ + , k∈ 24 4

b. ±

c. 1620°

π kπ + , 2 nπ, k, n ∈ 8 4

d.

kπ , k∈ 3

8. a.

e.

π kπ + , k∈ 18 6

f. π + 2 kπ,

b.

f.

10.

π 3π , 4 4

9. a. kπ , k∈ 2

12 + 2 kπ, k ∈ 5

d. −

π π + 2 kπ, + 2 arctan 2+ 2 n π, k, n ∈ 2 2

c.

π + kπ, − arctan 2+ nπ, k, n ∈ 4

i. 0

e. −

b.

e. ±

π 3. a. − + kπ, k ∈ 4

b.

π kπ + , k∈ 12 2

π + kπ, k ∈ 2

4. a.

π + kπ, k ∈ 4

f.

g.

b. ( −1)k

π kπ + , k∈ 4 2

h. c.

f. –2arccot 2 + 2kπ, k ∈

π kπ π + , ± + nπ, k, n ∈ 4 2 3

π 3 + kπ, ± arccos + 2 n π, k, n ∈ 2 4

π + 2 nπ, k, n ∈ 3

π 2 kπ + , k∈ 18 3

d. no solution

3π + 3kπ, k ∈ 4

π kπ π + , + nπ, k, n ∈ 8 4 2

e.

π kπ + , k∈ 30 5

π kπ, ( −1)n+1 + nπ, k, n ∈ 6

c.

π kπ n π + , , k, n ∈ 8 4 2

π kπ π + , ( −1)n +1 + nπ, k, n ∈ 4 2 6

π kπ + , k∈ 4 2

π 2 kπ 5 π 2 n π + , + , k, n ∈ 18 3 18 3

d.

b. ±

π kπ + , k∈ 6 2

π kπ + , k∈ 36 2

7π kπ + , k∈ 36 3

12. a. no solution

π 7π + 2kπ, + 2 nπ, k, n ∈ 12 12

e.

11.

f.

Answers to Exercises

π kπ + , k∈ 3 3

π + kπ, − arctan 2+ nπ, k, n ∈ 4

π + 2 nπ, k, n ∈ 2

π + 2 kπ, k ∈ 6

c. arctan

d.

7. a. kπ, ±

2π 4π , 3 3

π 3π 9π 11π , , , 8 8 8 8

d.

6. a.

π π + 2 kπ, ± + 2 nπ, k, n ∈ 3

π π 2 kπ + + , k∈ 20 20 5

e. –3arccot 2 + 3kπ, k ∈

3π kπ + , k∈ 8 2

c.

c. ±

h.

π kπ + , k∈ 90 3

d.

f.

d.

5π + 2 kπ, k ∈ 6

π 2kπ + , k∈ 30 5

g.

5. a. 4340 ° b. 3420°

c.

π or ( −1) k +1 + k π, k ∈ 3

d. ( −1)k

π 2. a. ± + 2kπ, k ∈ 4 π + kπ, k ∈ 6

π 2π b. − + 2kπ, − + 2 nπ, k, n ∈ 3 3

kπ 3π π kπ π π e. ( −1)k +1 + + kπ, k ∈ f. g. − +6 kπ, k ∈ + , k∈ , k∈ 9 3 3 5 5 4 2 ( −1)k +1 arcsin π 3 + kπ , k ∈ j. π + 4kπ, k ∈ i. k. ( −1)k +1 + 2 kπ, k ∈ l. no solution 4 4 2

π h. ± + kπ, k ∈ 3

f.

π or ( −1) k + k π, k ∈ 4

1 13. a. − arctan + kπ, k ∈ 2

π d. − + kπ,arctan3+ nπ, k, n ∈ 4

b.

e.

π + kπ, k ∈ 4

c.

π + kπ, k ∈ 4

4 b. − arctan + 2 kπ, k ∈ 3

π + 2kπ, π + 2 nπ, k, n ∈ 3

π + kπ, k ∈ 6

14. a. 5, –5 b. 2, –2

c. 3, –3

d. ñ2, –ñ2 487

15. a. [0, 26] b. [–2, 4] c. [–5, –1] π 3π , π, , 2π 2 2

18. 0,

22. ( x, y) = (2 kπ,

24.

19. kπ,

25. a. − d. −

π π 26. a. − + kπ < x < + kπ, k ∈ 3 3

π kπ π kπ b. − + < x< – + , k∈ 8 4 24 4

c. −

5π kπ π kπ + < x< + , k∈ 18 3 3 3

21. 0

7 π 2kπ π 2kπ + < x< + , k∈ 18 3 18 3

π 5π + kπ < x < + kπ, k ∈ 2 6

π + 4kπ < x < π + 4kπ, k ∈ 3

π π c. ( , 0), (0, ) 4 4

π π 23. ( −1)k +1 + kπ, ( −1)n + n π, k, n ∈ Z 3 3

17 π 2 kπ 8 π 2 kπ + + , k∈ ≤x≤ 45 3 45 3

b.

π b. ( , 0) 2

π π 17. a. ( , ) 6 6

20. –ò34 ≤ a ≤ ò34

π π − 2 kπ) or ( + k π, − k π), k ∈ Z 4 4

5π π c. − + kπ ≤ x ≤ + kπ, k ∈ 8 8

28. a.

π + kπ, nπ, k, n ∈ Z 2

π + 2 nπ, k, n ∈ Z 2

π + kπ, arccot2+ nπ, k, n ∈ Z 4

d. no solution e.

16.

π π b. − + 2kπ < x < + 2k π, k ∈ 2 6

e. x ≠ −

c. −

π 2 kπ + 10 5

9π 9π +6 kπ ≤ x ≤ +6 k π, k ∈ 4 4

π kπ π kπ 27. a. − + < x< + , k ∈ 6 3 12 3

2π 23 π + 2kπ ≤ x < + 2 k π, k ∈ 9 18

b. 2kπ < x ≤

d. −

π kπ π kπ + < x< + , k ∈ 15 5 10 5

3π + 2k π, k ∈ 2

π 7π ⎞ ⎛ 3π ⎡ π ⎤ 29. a. ⎢ − + 2kπ, + 2kπ ⎟ ∪ ⎜ + 2 nπ, + 2 nπ ⎥ , k, n ∈ 4 6 ⎣ 6 ⎦ ⎠ ⎝ 4 π π ⎛ π ⎞ ⎤ ⎡π b. ⎜ − + 2kπ, − + 2kπ ⎥ ∪ ⎢ + 2 nπ, + 2 nπ ⎟ , k, n ∈ 6 3 ⎦ ⎣6 ⎝ 3 ⎠ π ⎛ π ⎤ d. ⎜ − + kπ, + kπ⎥ , k ∈ 6 ⎝ 3 ⎦

e. ( kπ,

30. a. [–π + 2kπ, 2kπ], k ∈

b. (

c. ( −

2π π ⎡π ⎤ ⎡ 2π ⎤ c. ⎢ + 2 kπ, + 2 kπ ⎥ ∪ ⎢− + 2 n π, − + 2 n π ⎥, k, n ∈ 3 3 ⎣3 ⎦ ⎣ 3 ⎦

3π π + kπ) ∪ ( + k π, π+ k π), k, n ∈ 4 4

2π 4π + 2 kπ, + 2 kπ), k ∈ 3 3

2π 1 1 2π + 2 kπ, − arccos + 2 k π) ∪(arccos + 2 n π, + 2 n π), k, n ∈ 3 3 3 3

3π π e. ( + kπ, + kπ), k ∈ 4 4

π π 31 a. ( − + kπ, + kπ), k ∈ 3 3

π π π b. ( − + kπ, − arctan 2+ kπ) ∪( + n π, + n π), k, n ∈ 2 4 2 488

π ⎡ π ⎤ d. ⎢ − + kπ, + kπ ⎥ , k ∈ 4 4 ⎣ ⎦

5π π ⎡π ⎤ ⎡ 3π ⎤ c. ⎢ + kπ, + kπ ⎥ ∪ ⎢ + n π, + n π ⎥, k, n ∈ 6 4 4 6 ⎣ ⎦ ⎣ ⎦ Answers to Exercises

⎡ π 2kπ π 2kπ ⎤ 32. a. ⎢ − + , + , k∈ 3 12 3 ⎥⎦ ⎣ 12

d. ( −

7π π + kπ, + kπ), k ∈ 12 12

5π ⎡π ⎤ c. ⎢ + kπ, + kπ ⎥ , k ∈ 12 12 ⎣ ⎦

5π π e. ( + 2 kπ, + 2 kπ) ∪ ( π + 2 n π, 2 n π), k, n ∈ 6 6

π kπ π kπ + ,– + ), k ∈ 12 6 36 6 5π π c. ( + kπ, + kπ), k ∈ 3 12

π kπ π kπ b. ( + , + ), k ∈ 4 2 2 2

33. a. ( −

EXERCISES

⎡ 7π 2 kπ π 2 kπ ⎤ b. ⎢ − + , + , k∈ 3 18 3 ⎥⎦ ⎣ 18

34. a.

4 .1 1 1 5 4 g. – h. 2. a. y = 2x + 5 b. y = 3x c. y = 11x – 16 d. y = x + 2 3. –2 4 2 4 9

1. a. 5 b. –7 c. –2 d. –2 e. 0 f.

16 π 6. 9;11 7. a. rate of change of fuel consumption with respect to change in velocity b. if velocity 9

4. 30; 40 5.

8. a. x10; 1 b.

increases by 20 km/h, oil consumption decreases by 0.05 litres per hour 9. a. 2x – 2 b. –

3 ( x – 1)2

3

c. –

3

d.

2x x

2 3x +1

12. –2, 0, 6 : discontinuity; 1, 3 : corner 16. –2

17. (–2, 8), (2, 8)

EXERCISES 1. a. 0 b. 0 c. 0 d. m.

1 2 x

+

1 3

3 x

2

+

18. a. 50

b. 60

10. no; f ′(8–) ≠ f ′(8+) 11. does not exist

19. a = 1, b = 6, c = 0 20. 0

21. 9.95

1 – 2 7 1 x e. 1.6x–0.2 f. x 5 g. 0 h. 0.21x–0.3 i. –84 x–13 j. 10x – 3 k. 2 x + 2 + 2 l. –16t–5 + 9t–4 – 2t–2 3 x

1 5

5 x

n.

4

1 2 x

–

1

–

x x

l. − 3 3x + 2 x + 3 2 x(3x − 1)2

1 3 1 o. 2 – 2. a. 15x2 – 5 b. 12x + 1 c. –300x – 20 d. 4x3 + 3x2 – 1 x x2 2x x 3 – 21 t 2

g. –

6 (2 x + 4)2

h.

3 (2 x +1)2

3. a. 8 b. 2 c. –9 d. 1

4.

g. 3( x +1+ x )2 ⋅ (

1 2 x +1

+

1 2 x

)

4 h. 4( x − 1) (4 x +1)

(3x +1)

2 3

i. – 1

2 x

6. a. 6(3x – 1) b. 10x(x2 + 2)4 c. 7(x5– 3x2+6)6(5x4– 6x) d. –3(x–2)–4 e. −

Answers to Exercises

x; 8 c. 3x; 4 d. cos x; π

4 .2

3

x2 – 2 x – 2 ( x2 + x +1)2

3

13. no; discontinuity at x = 1 14. does not exist 15. D, E, C, A, B

e. 5x4 – 4x3 + 9x2 – 6x + 2 f. 5t 2 + 4t – k.

π π b. ( + kπ, + kπ), k ∈ 4 2

π + 2 kπ, k ∈ 4

6

5 (1+ 3x)2

–3

j.

1 − 3x2 2 x( x2 +1)2

5. a. –1 b. 8 c. –8 d. 8

40 x +12 4x f. − (5x2 + 3x − 1)3 (4 x2 +1)3

i. − (3x − 1) (18 x + 29) (2 x +1)5

2 j. 405( x − 3) 4 8( x + 2)

489

k. d. c.

1 – 5 2x − 1 3 2 3 2 3 2 l. 3 + [2 x + ( x +1) ] 4 [4 x + 2( x +1)3 x ] 4 2(2 x − 1)(3 x +1) 3 x+1

2

2( x – x) x – x 81 8(3x − 1)

3x − 1

2

d. 384x – 576 11. y = 7x – 5

12. – 2 x + 27 25 50

5 x x f ( x)+ x2 x f ′( x ) 2

13. a.

2 3 ′ b. 3x2[f(x)]2 + 2x3f(x)f ′(x) c. 3x f ( x) – x f ( x) d. 2 xf ( x)+ f ( x)+1 2 [ f ( x)] 2 x

16. a. C ′(t ) =

–3x2 ( x3 – 1)2

2 1 12 9. a. 6 b. 14(x2 + 1)5(13x2 + 1) c. ( x – 1)3 d. – 10. a. 120x – 18 b. − 4 (2 x – 1) 2 x – 1 x

(2 x – 1)( x2 – x – 1) 2

7. –12 8. a. 8x + 4 b. 2x c.

14. use the product rule twice 15. –

63 25

′ ′ ′ ⎛ ⎞ ′ 0.2 – 0.2 t2 b. 0, – 3 17. 6 18. f ′ ⎜ g( x)h( x) ⎟ ( g ( x)h( x)+ g( x)h ( x))m( n( x)) −2g( x)h( x)m ( n( x))n ( x) 2 2 (t + 1) 125 ( m( n( x))) ⎝ m( n( x)) ⎠

19. –3 + 4ñ3

21. − 15 4

20. 1

EXERCISES

22. ( –1)n

1 n! 2 xn+1

4 .3

1. a. 3cos(3x – 5) b. –2xsin(x2 – 1) c. cos x + sin x d. sec2x(2 + sin x) e. sin x(sec2x + 1) f. –cos x + 2tan x + 2xsec2x + xsin x g. −2 cos(2 x3 − 3 x)sin(2 x3 − 3 x) ⋅(6 x2 − 3) i. −

1 – cos x 9 2 sin x cos x(cos x + sin 2 x+1) ) j. 10( 2 2 1+ cos x (1+ cos x)2 sin x(cos x + 1)

m. 5( x2 sin( x − 1))4(2 x sin( x −1)+ x2 cos( x −1)) n. 3. ± 2π + 2π k, k ∈ 3 g.

4. a. 0 b.

b. –

3t ⋅ 3 t 2 4−t

2

6

4

2

– x + 3x – x

c.

490

1

π⋅ 2 4

8. a.

10. a.

2 x +1 1 x d. e. + 1+( x2 + x − 1)2 1+ x2 1 – x2

13. a. sin x b. –299cos 2x c. –35sinx – xcosx 15. y = 2x – 2 16.

2

20 sin x(1 − cos x) 9 (cos x +1)11

l. 9x2cot2(x3 – 1)csc2(x3 – 1)

2. y = –x

2x 6 x3 + y 2 3x2 + 2x − y 5 y 2 x +5 y c. 2 d. 0 5. 5 6. a. b. c. d. e. − f. − 3y x 3 x 4 1− x 5x + 3y2

4 17 f. – 5 cot t 9. a. y = x – 1 b. y = x – 3 3 4

x2 +1

x3 x3 tan 2 x −1 x −1 ( x2 − 1)2

(2 x4 − 6 x2 )sec 2

y – 2 xy3 y xy − 6xy h. 7. a. y = x – 3 b. y = x – x + 3x2 y2 4xy2 − x xy

e. −

2 2 k. (2 x − 1)sec x − x − 1 2 x2 − x − 1

h.

14. a.

t2 +1 2t – 2 2t + 3 b. 2 c. d. (4t +6) t +1 t (2t +1) 3 3t 2 – 2t 6( t2 − t − 1) 6 b. 2 c. 0 11. a. (2t − 1)3 t

1 4 − x2

1

12. y = 3x + π − 3 3 6 3 2 cos x − cos x 4

2

2 xy + 2y2 27 x4 3x 3 2 6y 3 b. c. d. 2 y + 2 y − − 3 7 3 2 ( x + 2 y) 16 y 2y x 3x 3 x 3x 3 x2

17. –sec2x 18. 48 19. (–∞, 0)\{–1} 20. 2 Answers to Exercises

EXERCISES 1. a. 0 b.

5 .1

6 5 1 3 1 5 3 1 1 2 1 c. d. –3 e. f. 3 g. – h. – i. − 2 ⋅ 3 j. k. cos a l. –1 m. – n. o. 6 2 2 6 4 3 5 48 4 9

b. –1 c. 1 d.

p.

1 2. a. 0 4

5 1 3 1 4 e. 3. a. b. 1 c. 0 d. 0 e. – f. 0 g. 3 3 2 π 3

EXERCISES

5 .2

1. a. increasing: (–∞, –2) and (0, 2); decreasing: (–2, 0) and (2, ∞) b. increasing: (–∞, –3); decreasing: (–3, 0) and (0, ∞) 2. a. decreasing: (–∞, ∞) b. increasing: (0, ∞); decreasing: (–∞, 0)

c. increasing: (–∞, 2); decreasing: (2, ∞)

d. increasing: (–∞, ∞) e. increasing: (–∞, 0) and (4, ∞); decreasing: (0, 4) f. increasing: (–2, 0) and (1, ∞); decreasing: (–∞, –2) and (0, 1) g. increasing: (–∞, 2) and (2, ∞) h. increasing: (0, ∞); decreasing: (–∞, 0) i. increasing: (–∞, 0) 5π π ⎞ and (10, ∞); decreasing: (0, 10) j. increasing: (–∞, ∞) k. increasing: ⎛⎜ − + 2π k, + 2π k ⎟, k ∈ ; decreasing: 6 ⎝ 6 ⎠ π 5π 1 7π ⎛π ⎞ ) 4. f ′( x) = > 0 for all x, so f is increasing 5. –2 < m < 2 6. 1 < a < 4 + 2π k ⎟ , k ∈ 3. ( , ⎜ + 2π k; 4 4 1+ x2 6 6 ⎝ ⎠

7. 1 < a < 3

8. –8

9. a. decreasing b. increasing c. decreasing d. increasing 10. a. decreasing b. increasing

c. increasing 11. local max.: x2, x4, x6; local min.: x3, x5, x7, x8; absolute max.: x6; absolute min.: x1, 12. a. no critical point b. –1 c. –1 13. a. f ′(a) = f ′(b) = f ′(c) = 0, f ′(d) and f ′(e) do not exist. b. local max.: a, d; local min.: b; neither: c, e 14. a. min.: x = –4 b. max.: x = 1; min.: x = –1 c. no local extrema d. min.: x = –3, x = 1; max.: x = –1 e. no local extrema f. min.: x = –2 g. max.: x = 0; min.: x = –2, x = 2 h. min.: x = 7π + 2π k, x = 11π + 2 π k; 6 6 max.: x =

π + π k, k ∈ 2

15. 6 16. a = 2, b = –4 17. k = 4 18. a = – 3 , b = –6 2

19. m = –3 20. ab < 0

21. –1 < m < 0 22. max.: x = –2, x = 7; min.: x = 3 23. a. f min =1, fmax = 3 b. f min = −9, fmax = 0 c. f min = −88, fmax = −7 d. f min =19, fmax = 30 e. f = −3, f = 9 − 4 3 f. f min = −255, fmax =1 min max g. f min = 0, fmax = 20

h. f = 5, f = 3 i. f min = 0, fmax = min max

l. f min = 0, fmax = 4 m. f min = −7, fmax = 3

8 3

j. f min = 0, fmax = 3

1 24. 24 25. 2 26. a = – , b = –3 4

k. f min = −1, fmax =

3 5

27. ñ2

28. a. increasing: (–2, ∞); decreasing: (–∞, –2) b. min.: x = –2 29. a. increasing: (–∞, –4) and (4, 6); decreasing: (–4, 4) and (6, ∞); b. min.: x = 4; max.: x = –4, x = 6 30. x = 2 32. (–∞, 6] ∪ [16, ∞) 33. a ≥ 3 Answers to Exercises

491

EXERCISES

5 .3

1. a. positive: (0, ∞); negative: (–∞, 0) b. positive: (–∞, –3), (–3, –1), (2, ∞); negative: (–1, 2) 2. a. concave up: (–∞, ∞), 1 1 no inflection point b. concave down: (–∞, ∞), no inflection point c. concave up: ( – , ∞ ), concave down: ( – ∞, – ), 3 3 inflection point x = – 1 d. concave up: (1, ∞), concave down: (–∞, 1), inflection point: x = 1 e. concave up: 3 5 5 5 (–∞, 1), ( , ∞ ), concave down: (1, ), inflection points: x = 1 and x = f. concave up: (–∞, –2), (2, ∞), 3 3 3 concave down: (–2, 2) inflection points: x = –2 and x = 2 g. concave up: (–1, 0), (1, ∞), concave down: (–∞, –1), (0, 1) inflection points: x = –1, x = 0, x = 1. h. concave up: (–∞, 0), (0, ∞), no inflection point i. concave down: (–1, ∞), no inflection point j. concave up: (5, ∞), concave down: (–∞, 5), inflection point: x = 5 k. concave up: (1, 2), concave down: (2, 3), inflection point x = 2 l. concave up: (2πk, π + 2πk), concave down: 3. a = 6, b = 0 4. (–1, –19) 5. concave up: (–1, 3), (π + 2πk, 2π + 2πk), inflection point x = π + πk, k ∈ concave down: (–∞, –1), (3, ∞), inflection points: x = –1, x = 3 6. a. minimum b. maximum c. minimum d. minimum 7. a. max.: x = –3, min.: x = –1 b. max : x = –3+ 5 , min : x = 0, x = –3 – 5 c. max.: x = 0, min.: x = 4 2 2 d. min.: x = –3 e. max.: x = 0 f. no extremum 8. a

9. consider the second derivative of f(x) = ax3 + bx2 + cx + d 10. consider the second

y b

derivative of f(x) = ax4 + bx3 + cx2 + dx + e 11. y = –x + 3 12. a = –1, b = 0, c = 3

x

13. c = 2 14. 2f ′(a) – f(a)

EXERCISES

5 .4

512 1. 15 and 15 2. –2 and 2 3. 50 4. 2ñm 5. m = 1 6. 50 m by 50 m 7. 30 m by 60 m 8. cm 3 9. 2, 2 and 1 m 3 5 6 10. 3888 cm3 11. 3 m 12. 250 13. 2R2 14. 2ñ2 15. (1, 0) 16. 24 17. y = 2ñ2 18. 13 19. r = 2 cm, h = cm 3 π 20. 1 hour 21. 9

EXERCISES 1. a. x = –1, y = 0 g. x = –3, y = 1 3. a.y

5 .5

b. x = 1, y = 0 h. x =

b.

–32

c.

y

3

6 x –3 –5/3

d. x = 4, y = –3

f. x = ±1, y = –1

e. x = –3, y = 0

3 3 i. x = 6, x = –1, y = 0 j. y = x k. x = 1, y = x + 3 , x = –3, y = – 2 2

y

4

c. no asymptotes

1 x

50 – 27

1 3

d. 1

e.

y –2

2

2

x

3

4 1 2. − , 9 4

f.

y

y

1

x –1

1

x

–1

1 x

–2 –12

492

Answers to Exercises

4. –

4 27

8 a.

5. a = –

3 7 , b= – 14 2

b.

y

c.

y

d.

y

e.

y

1

x

–1

1/3

x –1

1

x

–1

y

1

1/3 3

f.

y

1

1 –1

1 6. 0 7. y = ( x – 2)2 ( x +1)( x + 2) 2

1

5/3 3

–3

x

3

x

3 –3 2

x

–8

9 a.

b.

y 2 2

–2

c.

y

1

1

x

10. 1 solution if a ∈ (–∞, –4) ∪ (23, ∞), 2 solutions if a ∈ {–4, 23}, 3 solutions if a ∈ (–4, 23)

y

p/2

–2

2

–2

EXERCISES

6 .1

1. 12 2. 10 ways

3. 4 5. 12 6. 6 7. {MMM, MMF, MFM, MFF, FMM, FMF, FFM, FFF}

Windows 1 Doors

x

x

2

3

4

5

A

1A 2A 3A 4A 5A

B

1B 2B 3B 4B 5B

8. 12 9. 13 10. 4 11. 624 12. 46 13. 24 14. 6 15. 14 16. 412 17. a. 412 b. 512 18. 240 19. 28800 20. 264 = 456976 21. 5000 22. 4 000 000 23. a.10 b. 210 = 1028 24. 5850 25. 74360 26. No 27. No 28. a. 25 b. 10 29. 1944 30. 48 31. 135 32. 159 33. 8500 34. 12 35. a. 1792 b. 320 36. a. 140 b. 36 37. 9990000 38. 367 – 267 – 107 = 70322353920 39. 52(26 + 262 + 263) = 456950 40. Hint: Apply to the pigeonhole principle

EXERCISES

6 .2

⎛ Δ †  œ⎞ ⎟ 1. g−1 = ⎜ ⎜  Δ œ †⎟ ⎝ ⎠

⎛1 2 3 ⎞ ⎟ 2. ⎜ ⎜2 1 3⎟ ⎝ ⎠

9. 5! ⋅ 5! ⋅ 2! = 28800 10. a. 110

⎛1 2 3 4 ⎞ ⎟ 3. ⎜ ⎜4 3 2 1⎟ ⎝ ⎠

b. 81

c. n – 3

⎛a b c d e ⎞ ⎟ 4. ⎜ ⎜e c b a d⎟ ⎝ ⎠

d. 3

e.

1 42

5. 5! = 120

11. a. 2

b. 6

6. 720 c. 9

7. 120

8. 5040

d. 3 12. 336

13. P(10, 7) = 604800 14. P(10, 3) = 720 15. P(7, 3) = 210 16. P(7, 1) + P(7, 2) + P(7, 3) + P(7, 4) + P(7, 5) = 3619 17. P(26, 1) + P(26, 2) + P(26, 3) = 16276 18. 2! ⋅ 3! = 12

19. 64 – P(6, 4) = 576

22. 2 ⋅ P(26, 4) + P(26, 5) 26. a. 17!

23. 2 ⋅ 3! + 3 ⋅ 3! = 30

b. 3! ⋅ 4! ⋅ 6! ⋅ 7! = 522547200

Answers to Exercises

20. a. 120

b. 4 ⋅ P(6, 3) = 480 24. P(7, 4) ⋅ 24

27. 6! – 5! ⋅ 2! = 480

21. 4 ⋅ 5! = 480

25. 5! ⋅ 2! = 240

28. (2! ⋅ 3! ⋅ 5!) + (3! ⋅ 3! ⋅ 4!) = 2304 493

9! =1260 3! ⋅ 4! ⋅ 2! 20! 35. 9! ⋅ 2! ⋅ 2! ⋅ 2! ⋅ 2!⋅ 2!

29.

41. 3! ⋅ 5! = 720 47.

37. 126

c. 24

32.

44. 2! ⋅ 4! ⋅ 4! = 1152

48. 12 49. 360 50. 4 51.

EXERCISES b. 71

36. 60

42. 23! 43. 4! ⋅ 5! = 2880

8! =140 3! ⋅ 4! ⋅ 2!

1. a. 6

13! 33. 540 34. 50 = 60060 3! ⋅ 4! ⋅ 6! ⎛ 8! ⎞⎛ 4! ⎞ 12! 38. ⎜ b. 3! ⋅ 4! ⋅ 3! ⎟⎜ ⎟ = 168 39. a. ⋅ ⋅ 6! 2! 2! 2! 5! ⎝ ⎠⎝ ⎠

30. 7! = 5040 31. 75599

8! 4! ⋅ = 6 ⋅ 8! 2!⋅ 2! 2!

45. 5! ⋅ 3! = 720

46. 6! = 720

52. 21! ⋅ 222 53. 5 ⋅ 44 = 1280

d. 29 = 512

e. 212 = 4096

2. a. 12 b.

⎛ 16 ⎞ 7. 60 8. 35 9. 78 10. 45 11. 20 12. 10 13. ⎜ ⎟ =1820 ⎜4 ⎟ ⎝ ⎠

n−2 6

c. n2 – n

3. a. 6

⎛ 48 ⎞ 14. ⎜ ⎟ = 69668534468 ⎝ 12 ⎠

b. 9

⎛ 7 ⎞ ⎛ 3 ⎞⎛ 4 ⎞ ⎛ 3 ⎞⎛ 4 ⎞ ⎛ 3 ⎞⎛ 4 ⎞ 23. ⎜ ⎟ or ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ = 21 ⎝ 5 ⎠ ⎝ 1 ⎠⎝ 4 ⎠ ⎝ 2 ⎠⎝ 3 ⎠ ⎝ 3 ⎠⎝ 2 ⎠

⎛ 7 ⎞⎛ 5 ⎞ ⎛ 7 ⎞⎛ 5 ⎞ ⎛ 7 ⎞ b. ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ + ⎜ ⎟ = 246 ⎝ 5 ⎠⎝ 2 ⎠ ⎝ 6 ⎠⎝ 1 ⎠ ⎝ 7 ⎠

⎛ 19 ⎞ 39. a. ⎜ ⎟ = 3876 ⎝ 15 ⎠ ⎛5⎞ 43. ⎜ ⎟ ⋅ P(5,3) = 600 ⎝ 3⎠

⎛5⎞ ⎛ 5⎞ ⎛5⎞ ⎛5 ⎞ 26. ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ = 26 ⎝ 3⎠ ⎝ 2 ⎠ ⎝ 1⎠ ⎝0 ⎠

⎛8 ⎞ ⎛7 ⎞ 30. ⎜ ⎟ + ⎜ ⎟ + 3 = 94 ⎝ 3⎠ ⎝3⎠

⎛ 2010 ⎞ 35. ⎜ ⎟ = 678072034710 ⎝ 4 ⎠

⎛ 35 ⎞ b. ⎜ ⎟ = 3247943160 ⎝ 15 ⎠ ⎛7 ⎞ 44. ⎜ ⎟ ⋅ 4!= 504 ⎝5⎠

494

⎛6⎞ 40. 4 41. ⎜ ⎟ ⋅ 4! = 360 ⎝4⎠

48. 65 ⋅ 4! ⋅ 25 = 5971968

6. 10

⎛ 12 ⎞ ⎛10 ⎞ ⎛ 9 ⎞ 21. ⎜ ⎟ ⋅ ⎜ ⎟ ⋅ ⎜ ⎟ ⎝ 5 ⎠ ⎝ 4 ⎠ ⎝2⎠

⎛ 9⎞ ⎛ 3⎞ ⎛ 4 ⎞ 27. 21 28. a. 29 b. ⎜ ⎟ − ⎜ ⎟ − ⎜ ⎟ = 79 ⎝ 3⎠ ⎝ 3⎠ ⎝ 3 ⎠

⎛ 29 ⎞ ⎛ 37 ⎞ 37. ⎜ ⎟ ⋅ ⎜ ⎟ = 7843173975 ⎝5⎠ ⎝ 5⎠

⎛5⎞ ⎛6 ⎞ 45. a. ⎜ ⎟ ⋅ ⎜ ⎟ ⋅ 8! ⎝4⎠ ⎝4⎠

5. 64

⎛ 5 ⎞⎛ 7 ⎞ 25. a. ⎜ ⎟⎜ ⎟ = 350 ⎝ 3 ⎠⎝ 4 ⎠

⎛ 26 ⎞ 31. 45045 32. 600 33. ⎜ ⎟ = 14950 ⎝4 ⎠

⎛ 49 ⎞ 36. ⎜ ⎟ = 1906884 ⎝5 ⎠

⎛ 3 ⎞ ⎛ 6 ⎞ ⎛ 5 ⎞ ⎛ 18 ⎞ 47. ⎜ ⎟ ⋅ ⎜ ⎟ ⋅ ⎜ ⎟ ⋅ ⎜ ⎟ = 146880 ⎝ 1⎠ ⎝ 1⎠ ⎝ 1⎠ ⎝ 3 ⎠

51. 15

⎛8⎞ ⎛5⎞ 24. ⎜ ⎟ − ⎜ ⎟ = 65 ⎝4⎠ ⎝4⎠

c. 9

⎛ 40 ⎞ 15. ⎜ ⎟ = 91390 ⎝4 ⎠

⎛5⎞ ⎛5⎞ ⎛ 3 ⎞ ⎛ 3⎞ ⎛ 3⎞ ⎛ 12 ⎞ ⎛ 8 ⎞ ⎛ 11⎞ 16. ⎜ ⎟ = 330 17. ⎜ ⎟ + ⎜ ⎟ + ⎜ ⎟ = 9 18. ⎜ ⎟ ⋅ ⎜ ⎟ = 27720 19. 45 20. ⎜ ⎟ ⋅ ⎜ ⎟ ⋅ 3! = 600 ⎝2⎠ ⎝2⎠ ⎝ 2 ⎠ ⎝1 ⎠ ⎝1 ⎠ ⎝ 8 ⎠ ⎝3⎠ ⎝4⎠

⎛7 ⎞ ⎛ 4 ⎞ 29. ⎜ ⎟ ⋅ ⎜ ⎟ = 126 ⎝2⎠ ⎝2⎠

54. 2 ⋅ 18!

6 .3

4. {(k, l, m), (k, l, n), (k, l, r), (k, m, n), (k, m, r), (k, n, r), (l, m, n), (l, n, r), (l, m, r), (m, n, r)}

⎛7 ⎞ 22. ⎜ ⎟ = 21 ⎝2⎠

40. 4! = 24

⎛ 21⎞ 34. ⎜ ⎟ = 1330 ⎝3⎠ ⎛ 19 ⎞ 38. ⎜ ⎟ = 75582 ⎝ 11 ⎠

42. P(6, 4) = 360

⎛5⎞ b. P(5, 4) ⋅ P(6, 4) ⋅ 2! = 86400 46. ⎜ ⎟ ⋅ 34 = 810 ⎝2⎠ ⎛ 12 ⎞ 49. ⎜ ⎟ ⋅ 5! = 110880 ⎝6⎠

⎛7 ⎞ ⎛7 ⎞ 50. ⎜ ⎟ + ⎜ ⎟ = 56 ⎝ 3⎠ ⎝ 2 ⎠

⎛9⎞ 52. ⎜ ⎟ ⋅ 3! ⋅ 2! = 1008 ⎝ 3⎠ Answers to Exercises

EXERCISES

6 .4

1. a. 243x5 + 2025x4 + 6750x3 + 11250x2 + 9375x + 3125

b. 16x8 – 160x6 + 600x4 – 1000x2 + 625

c. 1024x15 – 3840x12y2 + 5760x9y4 – 4320x6y6 + 1620x3y8 – 243y10 ⎛ 15 ⎞ 5. ⎜ ⎟ (3x)4 ⎝ 11 ⎠

b. 515

4. a. 625

EXERCISES 1.

1 2

2.

2 9

3.

1 63

4.

1 4

5.

13.

3 4

1 60

6.

13 28

14.

7.

3 7

⎛ 210 ⎞⎛ 90 ⎞ ⎜ ⎟⎜ ⎟ ⎝ 1 ⎠⎝ 3 ⎠ = 9968 c. 133653 ⎛ 300 ⎞ ⎜ ⎟ ⎝ 4 ⎠

⎛ 4 ⎞⎛ 5 ⎞⎛ 4 ⎞ ⎛ 4 ⎞⎛ 5 ⎞ ⎛ 4 ⎞⎛ 4 ⎞ ⎜ ⎟⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ ⎝ 1 ⎠⎝ 1 ⎠⎝ 1 ⎠ ⎝ 1 ⎠⎝ 2 ⎠ ⎝ 1 ⎠⎝ 2 ⎠ = 72 21. 143 ⎛ 13 ⎞ ⎜ ⎟ ⎝3⎠

99 1 = ⎛ 100 ⎞ 50 ⎜ ⎟ ⎝ 2 ⎠

Answers to Exercises

⎛5⎞ ⎜ ⎟ ⎝ 4 ⎠ = 65 29. 1 − ⎛ 11⎞ 66 ⎜ ⎟ ⎝4⎠

8 11

8.

9. 3360

1 1 = ⎛ 42 ⎞ 111930 ⎜ ⎟ ⎝ 4⎠

⎛4⎞ ⎜ ⎟ 4 15. 1 − ⎝ ⎠ ⎛ 12 ⎞ ⎜ ⎟ ⎝4⎠ ⎛ 90 ⎞ ⎜ ⎟ 4 5162 d. ⎝ ⎠ = 300 ⎛ ⎞ 668265 ⎜ ⎟ 4 ⎝ ⎠

⎛ 10 ⎞ ⎜ ⎟ 3 67 22. 1 − ⎝ ⎠ = 15 ⎛ ⎞ 91 ⎜ ⎟ ⎝3⎠

⎛ 15 ⎞⎛ 5 ⎞ ⎛15 ⎞⎛ 5 ⎞ ⎛15 ⎞⎛ 5 ⎞ ⎛15 ⎞⎛ 5 ⎞ ⎛15 ⎞ ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ + ⎜ ⎟⎜ ⎟ + ⎜ ⎟ ⎝ 6 ⎠⎝ 4 ⎠ ⎝ 7 ⎠⎝ 3 ⎠ ⎝ 8 ⎠⎝ 2 ⎠ ⎝ 9 ⎠⎝1 ⎠ ⎝ 10 ⎠ 25. ⎛ 20 ⎞ ⎜ ⎟ ⎝ 10 ⎠

28.

8. 59

7. 240

10. 252

6 .5

⎛ 32 ⎞⎛168 ⎞ ⎜ ⎟⎜ ⎟ ⎝ 1 ⎠⎝ 3 ⎠ = 12419456 12. 32342475 ⎛ 200 ⎞ ⎜ ⎟ ⎝ 4 ⎠

⎛ 210 ⎞⎛ 90 ⎞ ⎜ ⎟⎜ ⎟ ⎜ 2 ⎟⎜ 2 ⎟ 11837 ⎝ ⎠⎝ ⎠ = b. 44551 ⎛ 300 ⎞ ⎜ ⎟ ⎜ 4 ⎟ ⎝ ⎠

⎛8⎞ 6. ⎜ ⎟ (3x4 )3 ⋅ ( y3 )5 ⎝5⎠

3. 600x2y2

2. –243

18.

31. 1 −

5! ⋅ 4! 9!

10. 150

11.

1 88 6

19.

7 18

⎛7 ⎞ ⎜ ⎟ 3 7 20. ⎝ ⎠ = ⎛ 10 ⎞ 24 ⎜ ⎟ ⎝3⎠

⎛ 7 ⎞⎛ 4 ⎞ ⎜ ⎟⎜ ⎟ 3 2 3 24. ⎝ ⎠⎝ ⎠ = 8 5 ⎛ ⎞⎛ ⎞ 10 ⎜ ⎟⎜ ⎟ ⎝ 4 ⎠⎝ 3 ⎠

⎛ 5 ⎞ ⎛ 5 ⎞⎛ 5 ⎞⎛ 5 ⎞ 3 ⋅ ⎜ ⎟ + ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ 3 ⎠ ⎝ 1 ⎠⎝ 1 ⎠⎝ 1 ⎠ = 31 27. 91 ⎛ 15 ⎞ ⎜ ⎟ ⎝3⎠

64 77 = ⎛ 9 ⎞⎛ 9 ⎞ 81 ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ 2 ⎠

1 63

⎛ 210 ⎞ ⎜ ⎟ ⎝ 4 ⎠ = 532 17. a. ⎛ 300 ⎞ 2235 ⎜ ⎟ ⎝ 4 ⎠

3 7

⎛7⎞ ⎜ ⎟ 3 1 23. 1 − ⎝ ⎠ = 8 ⎛ ⎞ 2 ⎜ ⎟ ⎝4⎠

⎛ 10 ⎞ ⎜ ⎟ ⋅ 6!⋅ 2! 1 ⎝5⎠ = 26. 11 ⎛ 12 ⎞ ⎜ ⎟ ⋅ 7! ⎝7⎠

⎛6⎞ 4 ⎜ ⎟⋅2 16 ⎝4⎠ = 30. 33 ⎛ 12 ⎞ ⎜ ⎟ ⎝4⎠

16.

9.

32.

50 1 = ⎛ 100 ⎞ 99 ⎜ ⎟ ⎝ 2 ⎠

495

TEST 1. 2. 3. 4. 5. 6. 7. 8.

A D E B D B D A

TEST 1. 2. 3. 4. 5. 6. 7. 8.

E B B A C A A D

TEST 1. 2. 3. 4. 5. 6. 7. 8.

C E C D A E C A

TEST 1. 2. 3. 4.

E D C D 496

1

TEST 9. 10. 11. 12. 13. 14. 15. 16.

C E C C B C D C

3C 9. 10. 11. 12. 13. 14. 15. 16.

A A E B C E E B

TEST C E E C A C B C

4A 9. 10. 11. 12. 13. 14. 15. 16.

1. 2. 3. 4. 5. 6. 7. 8.

1. 2. 3. 4. 5. 6. 7. 8.

D A E D B A D C

TEST B C C C B A A E

1. 2. 3. 4. 5. 6. 7. 8.

A E E C E C E B

2

TEST 9. 10. 11. 12. 13. 14. 15. 16.

A E C C E B E A

3D 9. 10. 11. 12. 13. 14. 15. 16.

E B A D D C B A

1. 2. 3. 4. 5. 6. 7. 8.

B D C D A C E D

TEST C C E E B D D B

6A 5. 6. 7. 8.

E A B E E E A E

TEST

4B 9. 10. 11. 12. 13. 14. 15. 16.

1. 2. 3. 4. 5. 6. 7. 8.

1. 2. 3. 4. 5. 6. 7. 8.

B C A E E E D A

TEST B A B E

9. 10. 11. 12.

D A C E

13. 14. 15. 16.

D A B C

1. 2. 3. 4.

A A E C

3A 9. 10. 11. 12. 13. 14. 15. 16.

TEST E C C D A B D C

3E 9. 10. 11. 12. 13. 14. 15. 16.

9. 10. 11. 12. 13. 14. 15. 16.

D B D B C D D E

TEST B C A A D E B C

5A 9. 10. 11. 12. 13. 14. 15. 16.

1. 2. 3. 4. 5. 6. 7. 8.

1. 2. 3. 4. 5. 6. 7. 8.

3B

3F 9. 10. 11. 12. 13. 14. 15.

B D C D E A C A

TEST

E B E C A D C D

D B C E B C A

5B

B C C D B A D E

1. 2. 3. 4. 5. 6. 7. 8.

E D B A E D E A

9. 10. 11. 12. 13. 14. 15. 16.

C D B E B A E B

A D C A

9. 10. 11. 12.

E B D D

13. 14. 15. 16.

E B D E

6B 5. 6. 7. 8.

Answers to Exercises

Answers to Exercises

497

498

Answers to Exercises

Answers to Exercises

499

500

Answers to Exercises